id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-011601 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Track quantifiers carefully: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{870} k\binom{870}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly expla... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{870\\cdot 2^{869}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{870\cdot 2^{869}$.) |
math-011602 | Combinatorics: Binomial Sums — Double Counting | 6 | Keep the final answer in boxed form: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1112} k\binom{1112}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Bri... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,1112\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1112\\cdot 2^{1111}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 111... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{1112\cdot 2^{1111}$.) |
math-011603 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Derive the result step-by-step: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2756} k\binom{2756}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ap... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,2756\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2756\\cdot 2^{2755}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{2756\cdot 2^{2755}$.) |
math-011604 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Give a theorem-based solution: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6879} k^2\binom{6879}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6879(6879+1)\\cdot 2^{6877}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6879(6879+1)\\cdot 2^{6877}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011605 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Track quantifiers carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{632} k^2\binom{632}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain careful... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{632(632+1)\\cdot 2^{630}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 632(632+1)\\cdot 2^{630}.",
"... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{632(632+1)\cdot 2^{630}$.) |
math-011606 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Answer with a short justification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6452} k^2\binom{6452}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain ca... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6452(6452+1)\\cdot 2^{6450}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6452(6452+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011607 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Find the exact value: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7722} k^2\binom{7722}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7722(7722+1)\\cdot 2^{7720}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7722(7722+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7722(7722+1)\cdot 2^{7720}$.) |
math-011608 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Solve (and briefly cross-validate): Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2369} k\binom{2369}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Brie... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2369\\cdot 2^{2368}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{2369\cdot 2^{2368}$.) |
math-011609 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Keep the final answer in boxed form: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1332} k^2\binom{1332}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tr... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1332\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1332(1332+1)\\cdot 2^{1330}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 1332(1332+1)\\cdot 2^{1330}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{1332(1332+1)\cdot 2^{1330}$.) |
math-011610 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Question: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7171} k^2\binom{7171}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefull... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7171(7171+1)\\cdot 2^{7169}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7171(7171+1)\\cdot 2^{7169}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7171(7171+1)\cdot 2^{7169}$.) |
math-011611 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Give an answer and a quick verification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6075} k\binom{6075}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c)... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,6075\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6075\\cdot 2^{6074}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{6075\cdot 2^{6074}$.) |
math-011612 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Give reasoning, not just computation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{444} k^2\binom{444}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain c... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,444\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{444(444+1)\\cdot 2^{442}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 444(444+1)\\cdot 2^{442}.",
"... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011613 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Make each step logically reversible (or explain if not): Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{486} k\binom{486}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argumen... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{486\\cdot 2^{485}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{486\cdot 2^{485}$.) |
math-011614 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Solve and justify each step: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3648} k\binom{3648}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly exp... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,3648\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3648\\cdot 2^{3647}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{3648\cdot 2^{3647}$.) |
math-011615 | Combinatorics: Binomial Sums — Double Counting | 6 | Prompt: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1839} k\binom{1839}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approac... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1839\\cdot 2^{1838}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011616 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Make each step logically reversible (or explain if not): Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1456} k^2\binom{1456}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate ... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1456(1456+1)\\cdot 2^{1454}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 1456(1456+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011617 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Give reasoning, not just computation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{374} k\binom{374}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why bot... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,374\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$,... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{374\\cdot 2^{373}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 374\\... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{374\cdot 2^{373}$.) |
math-011618 | Combinatorics: Binomial Sums — Double Counting | 6 | Proceed methodically: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7183} k\binom{7183}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both a... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7183\\cdot 2^{7182}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 718... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011619 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Carefully track domains: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2058} k^2\binom{2058}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully wh... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2058(2058+1)\\cdot 2^{2056}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2058(2058+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011620 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Give a theorem-based solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6862} k^2\binom{6862}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Exp... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6862(6862+1)\\cdot 2^{6860}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6862(6862+1)\\cdot 2^{6860}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{6862(6862+1)\cdot 2^{6860}$.) |
math-011621 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Track quantifiers carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5149} k\binom{5149}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ap... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5149\\cdot 2^{5148}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 514... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{5149\cdot 2^{5148}$.) |
math-011622 | Combinatorics: Binomial Sums — Double Counting | 6 | Answer with a short justification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6264} k^2\binom{6264}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of trip... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,6264\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6264(6264+1)\\cdot 2^{6262}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6264(6264+1)\\cdot 2^{6262}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{6264(6264+1)\cdot 2^{6262}$.) |
math-011623 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Complete the analysis: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2622} k^2\binom{2622}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully w... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2622\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2622(2622+1)\\cdot 2^{2620}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2622(2622+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011624 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Track quantifiers carefully: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6332} k^2\binom{6332}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6332(6332+1)\\cdot 2^{6330}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6332(6332+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{6332(6332+1)\cdot 2^{6330}$.) |
math-011625 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Derive the result step-by-step: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4684} k^2\binom{4684}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain caref... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4684(4684+1)\\cdot 2^{4682}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 4684(4684+1)\\cdot 2^{4682}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011626 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Start by stating any domain restrictions: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2396} k^2\binom{2396}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family ... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2396\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2396(2396+1)\\cdot 2^{2394}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2396(2396+1)\\cdot 2^{2394}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{2396(2396+1)\cdot 2^{2394}$.) |
math-011627 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Explain what is being counted/optimized: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2731} k\binom{2731}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain wh... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,2731\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2731\\cdot 2^{2730}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011628 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Where appropriate, name the theorem you use: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3841} k^2\binom{3841}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3841(3841+1)\\cdot 2^{3839}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 3841(3841+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011629 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Be explicit about assumptions: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5349} k^2\binom{5349}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain car... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5349(5349+1)\\cdot 2^{5347}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 5349(5349+1)\\cdot 2^{5347}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{5349(5349+1)\cdot 2^{5347}$.) |
math-011630 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Explain what is being counted/optimized: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4559} k\binom{4559}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c)... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4559\\cdot 2^{4558}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{4559\cdot 2^{4558}$.) |
math-011631 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Make each step logically reversible (or explain if not): Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{794} k^2\binom{794}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an approp... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{794(794+1)\\cdot 2^{792}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 794(794+1)\\cdot 2^{792}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{794(794+1)\cdot 2^{792}$.) |
math-011632 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Give reasoning, not just computation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6019} k^2\binom{6019}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of t... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6019(6019+1)\\cdot 2^{6017}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6019(6019+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{6019(6019+1)\cdot 2^{6017}$.) |
math-011633 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Be explicit about assumptions: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5179} k^2\binom{5179}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefu... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,5179\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5179(5179+1)\\cdot 2^{5177}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 5179(5179+1)\\cdot 2^{5177}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{5179(5179+1)\cdot 2^{5177}$.) |
math-011634 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Keep the final answer in boxed form: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7429} k^2\binom{7429}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expla... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,7429\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7429(7429+1)\\cdot 2^{7427}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7429(7429+1)\\cdot 2^{7427}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7429(7429+1)\cdot 2^{7427}$.) |
math-011635 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Make each step logically reversible (or explain if not): Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{233} k\binom{233}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Brie... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,233\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$,... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{233\\cdot 2^{232}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 233\\... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011636 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Give a theorem-based solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7310} k^2\binom{7310}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain car... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7310(7310+1)\\cdot 2^{7308}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7310(7310+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7310(7310+1)\cdot 2^{7308}$.) |
math-011637 | Combinatorics: Binomial Sums — Double Counting | 6 | Track units/moduli carefully: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7400} k\binom{7400}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly ex... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7400\\cdot 2^{7399}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 740... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011638 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Work carefully and justify each inference: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6187} k\binom{6187}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly expla... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,6187\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6187\\cdot 2^{6186}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{6187\cdot 2^{6186}$.) |
math-011639 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Give a fully justified solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2238} k^2\binom{2238}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain c... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2238\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2238(2238+1)\\cdot 2^{2236}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2238(2238+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{2238(2238+1)\cdot 2^{2236}$.) |
math-011640 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Determine the requested value: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4262} k^2\binom{4262}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4262\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4262(4262+1)\\cdot 2^{4260}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 4262(4262+1)\\cdot 2^{4260}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011641 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Proceed methodically: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7783} k^2\binom{7783}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain care... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7783(7783+1)\\cdot 2^{7781}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7783(7783+1)\\cdot 2^{7781}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7783(7783+1)\cdot 2^{7781}$.) |
math-011642 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Find the exact value: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2148} k^2\binom{2148}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2148(2148+1)\\cdot 2^{2146}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2148(2148+1)\\cdot 2^{2146}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{2148(2148+1)\cdot 2^{2146}$.) |
math-011643 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Write the solution set clearly: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{100} k\binom{100}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain wh... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,100\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$,... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{100\\cdot 2^{99}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{100\cdot 2^{99}$.) |
math-011644 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Determine the requested value: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1551} k^2\binom{1551}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1551\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1551(1551+1)\\cdot 2^{1549}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 1551(1551+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{1551(1551+1)\cdot 2^{1549}$.) |
math-011645 | Combinatorics: Binomial Sums — Double Counting | 6 | Solve with verification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6307} k\binom{6307}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approache... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,6307\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6307\\cdot 2^{6306}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011646 | Combinatorics: Binomial Sums — Double Counting | 6 | Answer using clear logical steps: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2021} k^2\binom{2021}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tripl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2021(2021+1)\\cdot 2^{2019}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2021(2021+1)\\cdot 2^{2019}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011647 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Explain what is being counted/optimized: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3014} k^2\binom{3014}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family o... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,3014\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3014(3014+1)\\cdot 2^{3012}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 3014(3014+1)\\cdot 2^{3012}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011648 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Give a fully justified solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{574} k\binom{574}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{574\\cdot 2^{573}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{574\cdot 2^{573}$.) |
math-011649 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Warm-up: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6418} k\binom{6418}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches cou... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6418\\cdot 2^{6417}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 641... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{6418\cdot 2^{6417}$.) |
math-011650 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Work carefully and justify each inference: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4871} k^2\binom{4871}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Ex... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4871\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4871(4871+1)\\cdot 2^{4869}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 4871(4871+1)\\cdot 2^{4869}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{4871(4871+1)\cdot 2^{4869}$.) |
math-011651 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Determine the requested value: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5045} k^2\binom{5045}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefu... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5045(5045+1)\\cdot 2^{5043}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 5045(5045+1)\\cdot 2^{5043}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011652 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Determine the requested value: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4861} k\binom{4861}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain w... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,4861\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4861\\cdot 2^{4860}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011653 | Combinatorics: Binomial Sums — Double Counting | 6 | Work carefully and justify each inference: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2241} k\binom{2241}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2241\\cdot 2^{2240}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{2241\cdot 2^{2240}$.) |
math-011654 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Show all reasoning: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2446} k^2\binom{2446}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why you... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2446(2446+1)\\cdot 2^{2444}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2446(2446+1)\\cdot 2^{2444}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{2446(2446+1)\cdot 2^{2444}$.) |
math-011655 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Show all reasoning: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7743} k^2\binom{7743}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefu... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7743(7743+1)\\cdot 2^{7741}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7743(7743+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7743(7743+1)\cdot 2^{7741}$.) |
math-011656 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Track units/moduli carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3821} k^2\binom{3821}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain care... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3821(3821+1)\\cdot 2^{3819}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 3821(3821+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{3821(3821+1)\cdot 2^{3819}$.) |
math-011657 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Solve with verification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2645} k^2\binom{2645}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2645(2645+1)\\cdot 2^{2643}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2645(2645+1)\\cdot 2^{2643}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011658 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Question: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3583} k^2\binom{3583}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why yo... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3583(3583+1)\\cdot 2^{3581}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 3583(3583+1)\\cdot 2^{3581}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011659 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Answer using clear logical steps: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3880} k\binom{3880}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explai... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,3880\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3880\\cdot 2^{3879}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 388... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{3880\cdot 2^{3879}$.) |
math-011660 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Checkpoint: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2405} k\binom{2405}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the s... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,2405\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2405\\cdot 2^{2404}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011661 | Combinatorics: Binomial Sums — Double Counting | 6 | Determine the requested value: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5346} k^2\binom{5346}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain car... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5346(5346+1)\\cdot 2^{5344}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 5346(5346+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{5346(5346+1)\cdot 2^{5344}$.) |
math-011662 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Track units/moduli carefully: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1625} k\binom{1625}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly ex... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,1625\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1625\\cdot 2^{1624}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011663 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Solve and include a self-check: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3455} k\binom{3455}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly ... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3455\\cdot 2^{3454}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{3455\cdot 2^{3454}$.) |
math-011664 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Make each step logically reversible (or explain if not): Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3391} k^2\binom{3391}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appr... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,3391\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3391(3391+1)\\cdot 2^{3389}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 3391(3391+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{3391(3391+1)\cdot 2^{3389}$.) |
math-011665 | Combinatorics: Binomial Sums — Double Counting | 6 | Track units/moduli carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7979} k^2\binom{7979}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain care... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7979(7979+1)\\cdot 2^{7977}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7979(7979+1)\\cdot 2^{7977}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7979(7979+1)\cdot 2^{7977}$.) |
math-011666 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Problem: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4555} k\binom{4555}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approa... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,4555\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4555\\cdot 2^{4554}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{4555\cdot 2^{4554}$.) |
math-011667 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Solve and sanity-check: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1937} k\binom{1937}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approac... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1937\\cdot 2^{1936}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{1937\cdot 2^{1936}$.) |
math-011668 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Where appropriate, name the theorem you use: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7353} k^2\binom{7353}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tr... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7353(7353+1)\\cdot 2^{7351}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7353(7353+1)\\cdot 2^{7351}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011669 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Answer with a short justification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6481} k^2\binom{6481}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain ca... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6481(6481+1)\\cdot 2^{6479}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6481(6481+1)\\cdot 2^{6479}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011670 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Provide both a computational and a conceptual explanation: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7754} k\binom{7754}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7754\\cdot 2^{7753}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 775... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{7754\cdot 2^{7753}$.) |
math-011671 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Answer using clear logical steps: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5860} k\binom{5860}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explai... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5860\\cdot 2^{5859}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{5860\cdot 2^{5859}$.) |
math-011672 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Solve with verification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2761} k^2\binom{2761}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully wh... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2761\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2761(2761+1)\\cdot 2^{2759}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 2761(2761+1)\\cdot 2^{2759}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{2761(2761+1)\cdot 2^{2759}$.) |
math-011673 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Solve and include a self-check: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7145} k^2\binom{7145}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain ca... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{7145(7145+1)\\cdot 2^{7143}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7145(7145+1)\\cdot 2^{7143}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7145(7145+1)\cdot 2^{7143}$.) |
math-011674 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Exercise: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5458} k^2\binom{5458}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why yo... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5458(5458+1)\\cdot 2^{5456}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 5458(5458+1)\\cdot 2^{5456}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{5458(5458+1)\cdot 2^{5456}$.) |
math-011675 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Explain why your operations are valid: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6773} k^2\binom{6773}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explai... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,6773\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6773(6773+1)\\cdot 2^{6771}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6773(6773+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{6773(6773+1)\cdot 2^{6771}$.) |
math-011676 | Combinatorics: Binomial Sums — Double Counting | 6 | Give reasoning, not just computation: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4188} k^2\binom{4188}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4188\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4188(4188+1)\\cdot 2^{4186}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 4188(4188+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{4188(4188+1)\cdot 2^{4186}$.) |
math-011677 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Answer using clear logical steps: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2037} k\binom{2037}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why bo... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2037\\cdot 2^{2036}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 203... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{2037\cdot 2^{2036}$.) |
math-011678 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Solve (and briefly cross-validate): Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7772} k\binom{7772}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why ... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,7772\\}$. Count pairs $(A,a)$ with $A\\subseteq[n]$ and $a\\in A$.",
"Step 2: If $|A|=k$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7772\\cdot 2^{7771}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011679 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Challenge: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{230} k^2\binom{230}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why your combinato... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,230\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{230(230+1)\\cdot 2^{228}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 230(230+1)\\cdot 2^{22... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{230(230+1)\cdot 2^{228}$.) |
math-011680 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Indicate where a theorem is used: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7792} k^2\binom{7792}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tripl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7792(7792+1)\\cdot 2^{7790}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7792(7792+1)\\cdot 2^{7790}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011681 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Do not skip justification steps: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7893} k^2\binom{7893}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain c... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,7893\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{7893(7893+1)\\cdot 2^{7891}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7893(7893+1)\\cdot 2^{7891}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7893(7893+1)\cdot 2^{7891}$.) |
math-011682 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Be explicit about assumptions: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4554} k^2\binom{4554}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefu... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4554(4554+1)\\cdot 2^{4552}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 4554(4554+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{4554(4554+1)\cdot 2^{4552}$.) |
math-011683 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Use two approaches if possible: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6446} k^2\binom{6446}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,6446\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6446(6446+1)\\cdot 2^{6444}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6446(6446+1)\\cdot 2^{6444}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011684 | Combinatorics: Binomial Sums — Double Counting | 6 | Complete the analysis: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1058} k^2\binom{1058}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Exp... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1058\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1058(1058+1)\\cdot 2^{1056}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 1058(1058+1)\\cdot 2^{1056}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011685 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Give an answer and a quick verification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4837} k^2\binom{4837}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family o... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4837(4837+1)\\cdot 2^{4835}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 4837(4837+1)\\cdot 2^{4835}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{4837(4837+1)\cdot 2^{4835}$.) |
math-011686 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Make each step logically reversible (or explain if not): Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5981} k\binom{5981}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c)... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5981\\cdot 2^{5980}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011687 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Determine the requested value: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4358} k^2\binom{4358}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4358\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4358(4358+1)\\cdot 2^{4356}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 4358(4358+1)\\cdot 2^{4356}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{4358(4358+1)\cdot 2^{4356}$.) |
math-011688 | Combinatorics: Binomial Sums — Double Counting | 6 | Problem: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6866} k^2\binom{6866}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6866(6866+1)\\cdot 2^{6864}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6866(6866+1)\\cdot 2^{6864}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{6866(6866+1)\cdot 2^{6864}$.) |
math-011689 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Derive the result step-by-step: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7921} k^2\binom{7921}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain ca... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{7921(7921+1)\\cdot 2^{7919}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7921(7921+1)\\cdot 2^{7919}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7921(7921+1)\cdot 2^{7919}$.) |
math-011690 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Compute the requested quantity: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1525} k^2\binom{1525}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Ex... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1525\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1525(1525+1)\\cdot 2^{1523}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 1525(1525+1)\\cdot 2^{1523}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{1525(1525+1)\cdot 2^{1523}$.) |
math-011691 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Answer with a short justification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6847} k\binom{6847}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why b... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6847\\cdot 2^{6846}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{6847\cdot 2^{6846}$.) |
math-011692 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Compute the requested quantity: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6784} k\binom{6784}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly ... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6784\\cdot 2^{6783}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 678... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{6784\cdot 2^{6783}$.) |
math-011693 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Do not skip justification steps: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7694} k\binom{7694}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why bot... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7694\\cdot 2^{7693}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011694 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 6 | Exercise: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3817} k\binom{3817}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both appro... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3817\\cdot 2^{3816}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-011695 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 6 | Give a fully justified solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3965} k^2\binom{3965}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain c... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3965(3965+1)\\cdot 2^{3963}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 3965(3965+1)\\cdot 2^{3963}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{3965(3965+1)\cdot 2^{3963}$.) |
math-011696 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Where appropriate, name the theorem you use: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{522} k\binom{522}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Brief... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{522\\cdot 2^{521}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 522\\... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{522\cdot 2^{521}$.) |
math-011697 | Combinatorics: Binomial Sums — Double Counting | 6 | Find the exact value: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6535} k^2\binom{6535}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why y... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,6535\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6535(6535+1)\\cdot 2^{6533}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6535(6535+1)\\cdot 2^{6533}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-011698 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 6 | Solve (and briefly cross-validate): Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7616} k^2\binom{7616}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explai... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7616(7616+1)\\cdot 2^{7614}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 7616(7616+1)\\cdot 2^{7614}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{7616(7616+1)\cdot 2^{7614}$.) |
math-011699 | Combinatorics: Binomial Sums — Double Counting | 6 | Show all reasoning: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5037} k\binom{5037}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches cou... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5037\\cdot 2^{5036}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 503... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{5037\cdot 2^{5036}$.) |
math-011700 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 6 | Answer using clear logical steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6266} k^2\binom{6266}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain car... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6266(6266+1)\\cdot 2^{6264}$.\nMethod (a) yields $S=n(n+1)2^{n-2}$. Method (b) counts the same triples $(A,a,b)$ and yields the identical closed form; here that closed form is 6266(6266+1)\\cdot 2^{6264}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
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