id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-015601 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Compute the requested quantity: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6158} k\binom{6158}{k}.$$
(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: $\\boxed{6158\\cdot 2^{6157}$.\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 615... | [
{
"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 $\boxed{6158\cdot 2^{6157}$.) |
math-015602 | Combinatorics: Binomial Sums — Double Counting | 8 | Explain what is being counted/optimized: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1502} k\binom{1502}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain wh... | [
{
"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: $\\boxed{1502\\cdot 2^{1501}$.\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 150... | [
{
"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 $\boxed{1502\cdot 2^{1501}$.) |
math-015603 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Solve (and briefly cross-validate): Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3500} k^2\binom{3500}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,3500\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3500(3500+1)\\cdot 2^{3498}$.\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 3500(3500+1)\\cdot 2^{3498}.... | [
{
"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-015604 | Combinatorics: Binomial Sums — Double Counting | 8 | Work carefully and justify each inference: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3782} k^2\binom{3782}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(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: $\\boxed{3782(3782+1)\\cdot 2^{3780}$.\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 3782(3782+1)\\cdot 2^{3780}.",
"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 $\boxed{3782(3782+1)\cdot 2^{3780}$.) |
math-015605 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Question: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2213} k^2\binom{2213}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,2213\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2213(2213+1)\\cdot 2^{2211}$.\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 2213(2213+1)\\cdot 2^{2211}.",
"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 $\boxed{2213(2213+1)\cdot 2^{2211}$.) |
math-015606 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Explain what is being counted/optimized: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1280} k^2\binom{1280}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1280(1280+1)\\cdot 2^{1278}$.\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 1280(1280+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-015607 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Track units/moduli carefully: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3442} k\binom{3442}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{3442\\cdot 2^{3441}$.\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 $\boxed{3442\cdot 2^{3441}$.) |
math-015608 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Answer with a short justification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5189} k\binom{5189}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{5189\\cdot 2^{5188}$.\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 518... | [
{
"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 $\boxed{5189\cdot 2^{5188}$.) |
math-015609 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Explain why your operations are valid: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6737} k\binom{6737}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why ... | [
{
"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: $\\boxed{6737\\cdot 2^{6736}$.\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 $(1+x)^n$. |
math-015610 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Give a fully justified solution: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6722} k^2\binom{6722}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,6722\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6722(6722+1)\\cdot 2^{6720}$.\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 6722(6722+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 $\boxed{6722(6722+1)\cdot 2^{6720}$.) |
math-015611 | Combinatorics: Binomial Sums — Double Counting | 8 | Try to avoid pattern-matching; explain why: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4252} k^2\binom{4252}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tri... | [
{
"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: $\\boxed{4252(4252+1)\\cdot 2^{4250}$.\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 4252(4252+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-015612 | Combinatorics: Binomial Sums — Double Counting | 8 | Track units/moduli carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3744} k^2\binom{3744}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,3744\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3744(3744+1)\\cdot 2^{3742}$.\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 3744(3744+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 $\boxed{3744(3744+1)\cdot 2^{3742}$.) |
math-015613 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Start by stating any domain restrictions: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5484} k\binom{5484}{k}.$$
(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 $[n]=\\{1,2,\\dots,5484\\}$. 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: $\\boxed{5484\\cdot 2^{5483}$.\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 $\boxed{5484\cdot 2^{5483}$.) |
math-015614 | Combinatorics: Binomial Sums — Double Counting | 8 | Solve and justify each step: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6984} k^2\binom{6984}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{6984(6984+1)\\cdot 2^{6982}$.\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 6984(6984+1)\\cdot 2^{6982}.",
"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 $\boxed{6984(6984+1)\cdot 2^{6982}$.) |
math-015615 | Combinatorics: Binomial Sums — Double Counting | 8 | Prompt: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4475} k^2\binom{4475}{k}.$$
(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 combina... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,4475\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4475(4475+1)\\cdot 2^{4473}$.\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 4475(4475+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 $\boxed{4475(4475+1)\cdot 2^{4473}$.) |
math-015616 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Give a fully justified solution: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4784} k^2\binom{4784}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triple... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,4784\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4784(4784+1)\\cdot 2^{4782}$.\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 4784(4784+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 $\boxed{4784(4784+1)\cdot 2^{4782}$.) |
math-015617 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | State any required conditions first: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2265} k^2\binom{2265}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2265\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2265(2265+1)\\cdot 2^{2263}$.\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 2265(2265+1)\\cdot 2^{2263}.... | [
{
"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 $\boxed{2265(2265+1)\cdot 2^{2263}$.) |
math-015618 | Combinatorics: Binomial Sums — Double Counting | 8 | Write the solution set clearly: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7956} k^2\binom{7956}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,7956\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7956(7956+1)\\cdot 2^{7954}$.\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 7956(7956+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 $\boxed{7956(7956+1)\cdot 2^{7954}$.) |
math-015619 | Combinatorics: Binomial Sums — Double Counting | 8 | Start by stating any domain restrictions: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6850} k^2\binom{6850}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6850\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6850(6850+1)\\cdot 2^{6848}$.\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 6850(6850+1)\\cdot 2^{6848}.... | [
{
"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 $\boxed{6850(6850+1)\cdot 2^{6848}$.) |
math-015620 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Give an answer and a quick verification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6027} k\binom{6027}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain... | [
{
"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: $\\boxed{6027\\cdot 2^{6026}$.\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 602... | [
{
"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-015621 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Find the exact value: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7449} k\binom{7449}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both a... | [
{
"method_name": "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 $[n]=\\{1,2,\\dots,7449\\}$. 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: $\\boxed{7449\\cdot 2^{7448}$.\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 744... | [
{
"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-015622 | Combinatorics: Binomial Sums — Double Counting | 8 | Provide both a computational and a conceptual explanation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2721} k^2\binom{2721}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of... | [
{
"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: $\\boxed{2721(2721+1)\\cdot 2^{2719}$.\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 2721(2721+1)\\cdot 2^{2719}.... | [
{
"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 $\boxed{2721(2721+1)\cdot 2^{2719}$.) |
math-015623 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Start by stating any domain restrictions: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2527} k\binom{2527}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{2527\\cdot 2^{2526}$.\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 $\boxed{2527\cdot 2^{2526}$.) |
math-015624 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | State any required conditions first: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5102} k\binom{5102}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5102\\cdot 2^{5101}$.\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 $\boxed{5102\cdot 2^{5101}$.) |
math-015625 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Be explicit about assumptions: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4731} k\binom{4731}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly e... | [
{
"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: $\\boxed{4731\\cdot 2^{4730}$.\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 $\boxed{4731\cdot 2^{4730}$.) |
math-015626 | Combinatorics: Binomial Sums — Double Counting | 8 | Explain each transformation: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5905} k\binom{5905}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5905\\cdot 2^{5904}$.\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 $\boxed{5905\cdot 2^{5904}$.) |
math-015627 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Complete the analysis: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2054} k^2\binom{2054}{k}.$$
(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 ... | [
{
"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: $\\boxed{2054(2054+1)\\cdot 2^{2052}$.\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 2054(2054+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 $\boxed{2054(2054+1)\cdot 2^{2052}$.) |
math-015628 | Combinatorics: Binomial Sums — Double Counting | 8 | Give an answer and a quick verification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1858} k^2\binom{1858}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,1858\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1858(1858+1)\\cdot 2^{1856}$.\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 1858(1858+1)\\cdot 2^{1856}.",
"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 $k$. |
math-015629 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Answer with a short justification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4939} k\binom{4939}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why b... | [
{
"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 $[n]=\\{1,2,\\dots,4939\\}$. 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: $\\boxed{4939\\cdot 2^{4938}$.\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 493... | [
{
"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 $\boxed{4939\cdot 2^{4938}$.) |
math-015630 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Checkpoint: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7181} k^2\binom{7181}{k}.$$
(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 com... | [
{
"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: $\\boxed{7181(7181+1)\\cdot 2^{7179}$.\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 7181(7181+1)\\cdot 2^{7179}.... | [
{
"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-015631 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Proceed methodically: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1759} k\binom{1759}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches 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: $\\boxed{1759\\cdot 2^{1758}$.\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... | 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-015632 | Combinatorics: Binomial Sums — Double Counting | 8 | Answer using clear logical steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1998} k\binom{1998}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ... | [
{
"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 $[n]=\\{1,2,\\dots,1998\\}$. 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: $\\boxed{1998\\cdot 2^{1997}$.\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 $(1+x)^n$. (Here the result is $\boxed{1998\cdot 2^{1997}$.) |
math-015633 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Solve and justify each step: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4260} k^2\binom{4260}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4260(4260+1)\\cdot 2^{4258}$.\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 4260(4260+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-015634 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Give an answer and a quick verification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3158} k^2\binom{3158}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triple... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,3158\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3158(3158+1)\\cdot 2^{3156}$.\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 3158(3158+1)\\cdot 2^{3156}.... | [
{
"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-015635 | Combinatorics: Binomial Sums — Double Counting | 8 | Explain each transformation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3505} k\binom{3505}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both appro... | [
{
"method_name": "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 $[n]=\\{1,2,\\dots,3505\\}$. 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: $\\boxed{3505\\cdot 2^{3504}$.\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 350... | [
{
"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 $\boxed{3505\cdot 2^{3504}$.) |
math-015636 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Give a theorem-based solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5040} k\binom{5040}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{5040\\cdot 2^{5039}$.\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 $(1+x)^n$. (Here the result is $\boxed{5040\cdot 2^{5039}$.) |
math-015637 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Do not skip justification steps: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4742} k^2\binom{4742}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) E... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,4742\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4742(4742+1)\\cdot 2^{4740}$.\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 4742(4742+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-015638 | Combinatorics: Binomial Sums — Double Counting | 8 | Complete the analysis: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6152} k\binom{6152}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6152\\cdot 2^{6151}$.\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 $(1+x)^n$. (Here the result is $\boxed{6152\cdot 2^{6151}$.) |
math-015639 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Give an answer and a quick verification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5990} k^2\binom{5990}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triple... | [
{
"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: $\\boxed{5990(5990+1)\\cdot 2^{5988}$.\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 5990(5990+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-015640 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Use two approaches if possible: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1835} k\binom{1835}{k}.$$
(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 $[n]=\\{1,2,\\dots,1835\\}$. 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: $\\boxed{1835\\cdot 2^{1834}$.\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 $(1+x)^n$. |
math-015641 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Provide both a computational and a conceptual explanation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7244} k\binom{7244}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{7244\\cdot 2^{7243}$.\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 724... | [
{
"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-015642 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Solve with verification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1086} k^2\binom{1086}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{1086(1086+1)\\cdot 2^{1084}$.\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 1086(1086+1)\\cdot 2^{1084}.",
"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 $\boxed{1086(1086+1)\cdot 2^{1084}$.) |
math-015643 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Answer using clear logical steps: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5810} k\binom{5810}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5810\\cdot 2^{5809}$.\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 $\boxed{5810\cdot 2^{5809}$.) |
math-015644 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Indicate where a theorem is used: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2632} k\binom{2632}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why bo... | [
{
"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 $[n]=\\{1,2,\\dots,2632\\}$. 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: $\\boxed{2632\\cdot 2^{2631}$.\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-015645 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Try to avoid pattern-matching; explain why: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6476} k^2\binom{6476}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tri... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,6476\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6476(6476+1)\\cdot 2^{6474}$.\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 6476(6476+1)\\cdot 2^{6474}.",
"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 $\boxed{6476(6476+1)\cdot 2^{6474}$.) |
math-015646 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Answer with a short justification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{427} k\binom{427}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both a... | [
{
"method_name": "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 $[n]=\\{1,2,\\dots,427\\}$. 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: $\\boxed{427\\cdot 2^{426}$.\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 $(1+x)^n$. |
math-015647 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Provide a rigorous solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4625} k^2\binom{4625}{k}.$$
(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": "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: $\\boxed{4625(4625+1)\\cdot 2^{4623}$.\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 4625(4625+1)\\cdot 2^{4623}.",
"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 $k$. |
math-015648 | Combinatorics: Binomial Sums — Double Counting | 8 | Proceed methodically: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7019} k\binom{7019}{k}.$$
(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 $[n]=\\{1,2,\\dots,7019\\}$. 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: $\\boxed{7019\\cdot 2^{7018}$.\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-015649 | Combinatorics: Binomial Sums — Double Counting | 8 | Make each step logically reversible (or explain if not): Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2267} k\binom{2267}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Br... | [
{
"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: $\\boxed{2267\\cdot 2^{2266}$.\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 $\boxed{2267\cdot 2^{2266}$.) |
math-015650 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Try to avoid pattern-matching; explain why: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1670} k^2\binom{1670}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate famil... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,1670\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1670(1670+1)\\cdot 2^{1668}$.\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 1670(1670+1)\\cdot 2^{1668}.",
"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 $\boxed{1670(1670+1)\cdot 2^{1668}$.) |
math-015651 | Combinatorics: Binomial Sums — Double Counting | 8 | Derive the result step-by-step: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4021} k^2\binom{4021}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4021(4021+1)\\cdot 2^{4019}$.\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 4021(4021+1)\\cdot 2^{4019}.... | [
{
"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-015652 | Combinatorics: Binomial Sums — Double Counting | 8 | Explain what is being counted/optimized: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5136} k\binom{5136}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{5136\\cdot 2^{5135}$.\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 513... | [
{
"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-015653 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Solve and sanity-check: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1198} k^2\binom{1198}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,1198\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1198(1198+1)\\cdot 2^{1196}$.\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 1198(1198+1)\\cdot 2^{1196}.... | [
{
"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 $\boxed{1198(1198+1)\cdot 2^{1196}$.) |
math-015654 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Keep the final answer in boxed form: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{338} k^2\binom{338}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{338(338+1)\\cdot 2^{336}$.\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 338(338+1)\\cdot 2^{336}.",
"... | [
{
"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-015655 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Use two approaches if possible: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{873} k\binom{873}{k}.$$
(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 $[n]=\\{1,2,\\dots,873\\}$. 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: $\\boxed{873\\cdot 2^{872}$.\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... | 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 $\boxed{873\cdot 2^{872}$.) |
math-015656 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | State any required conditions first: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4134} k^2\binom{4134}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain ... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,4134\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4134(4134+1)\\cdot 2^{4132}$.\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 4134(4134+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 $\boxed{4134(4134+1)\cdot 2^{4132}$.) |
math-015657 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Track units/moduli carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2531} k\binom{2531}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{2531\\cdot 2^{2530}$.\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-015658 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Explain each transformation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6391} k^2\binom{6391}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,6391\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6391(6391+1)\\cdot 2^{6389}$.\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 6391(6391+1)\\cdot 2^{6389}.",
"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 $\boxed{6391(6391+1)\cdot 2^{6389}$.) |
math-015659 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Track quantifiers carefully: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6587} k^2\binom{6587}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6587\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6587(6587+1)\\cdot 2^{6585}$.\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 6587(6587+1)\\cdot 2^{6585}.",
"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 $\boxed{6587(6587+1)\cdot 2^{6585}$.) |
math-015660 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | 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
$$S=\sum_{k=0}^{6720} k\binom{6720}{k}.$$
(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 $[n]=\\{1,2,\\dots,6720\\}$. 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: $\\boxed{6720\\cdot 2^{6719}$.\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 672... | [
{
"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-015661 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Solve with verification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4198} k^2\binom{4198}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,4198\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4198(4198+1)\\cdot 2^{4196}$.\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 4198(4198+1)\\cdot 2^{4196}.... | [
{
"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 $\boxed{4198(4198+1)\cdot 2^{4196}$.) |
math-015662 | Combinatorics: Binomial Sums — Double Counting | 8 | Give an answer and a quick verification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4109} k\binom{4109}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain... | [
{
"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: $\\boxed{4109\\cdot 2^{4108}$.\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 410... | [
{
"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 $\boxed{4109\cdot 2^{4108}$.) |
math-015663 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | 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
$$S=\sum_{k=0}^{127} k^2\binom{127}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{127(127+1)\\cdot 2^{125}$.\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 127(127+1)\\cdot 2^{12... | [
{
"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 $\boxed{127(127+1)\cdot 2^{125}$.) |
math-015664 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Work carefully and justify each inference: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3197} k^2\binom{3197}{k}.$$
(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": "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: $\\boxed{3197(3197+1)\\cdot 2^{3195}$.\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 3197(3197+1)\\cdot 2^{3195}.",
"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... | 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 $\boxed{3197(3197+1)\cdot 2^{3195}$.) |
math-015665 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Compute the requested quantity: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6369} k^2\binom{6369}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6369\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6369(6369+1)\\cdot 2^{6367}$.\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 6369(6369+1)\\cdot 2^{6367}.",
"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... | 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-015666 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Give a fully justified solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2656} k^2\binom{2656}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) E... | [
{
"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: $\\boxed{2656(2656+1)\\cdot 2^{2654}$.\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 2656(2656+1)\\cdot 2^{2654}.... | [
{
"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-015667 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Show all reasoning: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1479} k^2\binom{1479}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1479(1479+1)\\cdot 2^{1477}$.\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 1479(1479+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 $\boxed{1479(1479+1)\cdot 2^{1477}$.) |
math-015668 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Write the solution set clearly: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4172} k^2\binom{4172}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,4172\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4172(4172+1)\\cdot 2^{4170}$.\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 4172(4172+1)\\cdot 2^{4170}.",
"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... | 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-015669 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Question: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7401} k^2\binom{7401}{k}.$$
(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 combinat... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,7401\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7401(7401+1)\\cdot 2^{7399}$.\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 7401(7401+1)\\cdot 2^{7399}.",
"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 $k$. |
math-015670 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Solve and sanity-check: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5408} k\binom{5408}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches... | [
{
"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: $\\boxed{5408\\cdot 2^{5407}$.\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 $\boxed{5408\cdot 2^{5407}$.) |
math-015671 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | Derive the result step-by-step: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2735} k^2\binom{2735}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,2735\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2735(2735+1)\\cdot 2^{2733}$.\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 2735(2735+1)\\cdot 2^{2733}.... | [
{
"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-015672 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Do not skip justification steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1344} k\binom{1344}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1344\\cdot 2^{1343}$.\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-015673 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Try to avoid pattern-matching; explain why: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3049} k\binom{3049}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
... | [
{
"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 $[n]=\\{1,2,\\dots,3049\\}$. 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: $\\boxed{3049\\cdot 2^{3048}$.\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 $(1+x)^n$. (Here the result is $\boxed{3049\cdot 2^{3048}$.) |
math-015674 | Combinatorics: Binomial Sums — Double Counting | 8 | Solve (and briefly cross-validate): Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3105} k\binom{3105}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why ... | [
{
"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: $\\boxed{3105\\cdot 2^{3104}$.\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... | 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-015675 | Combinatorics: Binomial Sums — Double Counting | 8 | Solve with verification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4371} k^2\binom{4371}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4371(4371+1)\\cdot 2^{4369}$.\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 4371(4371+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 $\boxed{4371(4371+1)\cdot 2^{4369}$.) |
math-015676 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Provide a rigorous solution: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{844} k^2\binom{844}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,844\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{844(844+1)\\cdot 2^{842}$.\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 844(844+1)\\cdot 2^{842}.",
"... | [
{
"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 $\boxed{844(844+1)\cdot 2^{842}$.) |
math-015677 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Find the exact value: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4609} k^2\binom{4609}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,4609\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4609(4609+1)\\cdot 2^{4607}$.\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 4609(4609+1)\\cdot 2^{4607}.",
"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 $\boxed{4609(4609+1)\cdot 2^{4607}$.) |
math-015678 | Combinatorics: Binomial Sums — Double Counting | 8 | Work carefully and justify each inference: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2741} k^2\binom{2741}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2741\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2741(2741+1)\\cdot 2^{2739}$.\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 2741(2741+1)\\cdot 2^{2739}.",
"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-015679 | Combinatorics: Binomial Sums — Double Counting | 8 | Carefully track domains: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{918} k^2\binom{918}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,918\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{918(918+1)\\cdot 2^{916}$.\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 918(918+1)\\cdot 2^{916}.",
"... | [
{
"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 $\boxed{918(918+1)\cdot 2^{916}$.) |
math-015680 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Exercise: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1372} k^2\binom{1372}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{1372(1372+1)\\cdot 2^{1370}$.\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 1372(1372+1)\\cdot 2^{1370}.",
"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 $\boxed{1372(1372+1)\cdot 2^{1370}$.) |
math-015681 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Task: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4329} k\binom{4329}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count ... | [
{
"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 $[n]=\\{1,2,\\dots,4329\\}$. 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: $\\boxed{4329\\cdot 2^{4328}$.\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 $\boxed{4329\cdot 2^{4328}$.) |
math-015682 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Compute the requested quantity: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3327} k^2\binom{3327}{k}.$$
(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": "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: $\\boxed{3327(3327+1)\\cdot 2^{3325}$.\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 3327(3327+1)\\cdot 2^{3325}.",
"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 $\boxed{3327(3327+1)\cdot 2^{3325}$.) |
math-015683 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | State any required conditions first: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6285} k^2\binom{6285}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6285\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6285(6285+1)\\cdot 2^{6283}$.\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 6285(6285+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 $\boxed{6285(6285+1)\cdot 2^{6283}$.) |
math-015684 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Write the solution set clearly: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6295} k^2\binom{6295}{k}.$$
(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: $\\boxed{6295(6295+1)\\cdot 2^{6293}$.\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 6295(6295+1)\\cdot 2^{6293}.",
"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... | 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 $\boxed{6295(6295+1)\cdot 2^{6293}$.) |
math-015685 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Solve and sanity-check: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7752} k^2\binom{7752}{k}.$$
(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: $\\boxed{7752(7752+1)\\cdot 2^{7750}$.\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 7752(7752+1)\\cdot 2^{7750}.",
"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-015686 | Combinatorics: Binomial Sums — Double Counting | 8 | Do not skip justification steps: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6106} k^2\binom{6106}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{6106(6106+1)\\cdot 2^{6104}$.\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 6106(6106+1)\\cdot 2^{6104}.",
"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 $k$. |
math-015687 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | 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
$$S=\sum_{k=0}^{4062} k\binom{4062}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Bri... | [
{
"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: $\\boxed{4062\\cdot 2^{4061}$.\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... | 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 $\boxed{4062\cdot 2^{4061}$.) |
math-015688 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Question: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2481} k\binom{2481}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the sam... | [
{
"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: $\\boxed{2481\\cdot 2^{2480}$.\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 $(1+x)^n$. (Here the result is $\boxed{2481\cdot 2^{2480}$.) |
math-015689 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 8 | Keep the final answer in boxed form: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6370} k\binom{6370}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Bri... | [
{
"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: $\\boxed{6370\\cdot 2^{6369}$.\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 $(1+x)^n$. (Here the result is $\boxed{6370\cdot 2^{6369}$.) |
math-015690 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Provide a rigorous solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6754} k^2\binom{6754}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6754\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6754(6754+1)\\cdot 2^{6752}$.\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 6754(6754+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-015691 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Provide both a computational and a conceptual explanation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6434} k^2\binom{6434}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,6434\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6434(6434+1)\\cdot 2^{6432}$.\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 6434(6434+1)\\cdot 2^{6432}.",
"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 $\boxed{6434(6434+1)\cdot 2^{6432}$.) |
math-015692 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | Carefully track domains: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1090} k\binom{1090}{k}.$$
(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 $[n]=\\{1,2,\\dots,1090\\}$. 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: $\\boxed{1090\\cdot 2^{1089}$.\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 109... | [
{
"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 $\boxed{1090\cdot 2^{1089}$.) |
math-015693 | Combinatorics: Binomial Sums — Double Counting | 8 | Prompt: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3491} k^2\binom{3491}{k}.$$
(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 combinator... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,3491\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3491(3491+1)\\cdot 2^{3489}$.\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 3491(3491+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 $\boxed{3491(3491+1)\cdot 2^{3489}$.) |
math-015694 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 8 | 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
$$S=\sum_{k=0}^{476} k^2\binom{476}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an approp... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,476\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{476(476+1)\\cdot 2^{474}$.\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 476(476+1)\\cdot 2^{474}.",
"robustness_analysis": "... | [
{
"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 $\boxed{476(476+1)\cdot 2^{474}$.) |
math-015695 | Combinatorics: Binomial Sums — Double Counting | 8 | Compute the requested quantity: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1740} k\binom{1740}{k}.$$
(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 $[n]=\\{1,2,\\dots,1740\\}$. 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: $\\boxed{1740\\cdot 2^{1739}$.\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... | 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 $\boxed{1740\cdot 2^{1739}$.) |
math-015696 | Combinatorics: Binomial Sums — Double Counting | 8 | Challenge: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3987} k^2\binom{3987}{k}.$$
(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 comb... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,3987\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3987(3987+1)\\cdot 2^{3985}$.\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 3987(3987+1)\\cdot 2^{3985}.",
"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 $\boxed{3987(3987+1)\cdot 2^{3985}$.) |
math-015697 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Answer using clear logical steps: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3607} k\binom{3607}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{3607\\cdot 2^{3606}$.\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 $\boxed{3607\cdot 2^{3606}$.) |
math-015698 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 8 | Provide a rigorous solution: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3654} k^2\binom{3654}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{3654(3654+1)\\cdot 2^{3652}$.\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 3654(3654+1)\\cdot 2^{3652}.",
"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 $\boxed{3654(3654+1)\cdot 2^{3652}$.) |
math-015699 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 8 | 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
$$S=\sum_{k=0}^{2951} k^2\binom{2951}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2951\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2951(2951+1)\\cdot 2^{2949}$.\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 2951(2951+1)\\cdot 2^{2949}.... | [
{
"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 $\boxed{2951(2951+1)\cdot 2^{2949}$.) |
math-015700 | Combinatorics: Binomial Sums — Double Counting | 8 | Answer with a short justification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2101} k\binom{2101}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both... | [
{
"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 $[n]=\\{1,2,\\dots,2101\\}$. 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: $\\boxed{2101\\cdot 2^{2100}$.\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 $\boxed{2101\cdot 2^{2100}$.) |
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