id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-013601 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Explain each transformation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{272} k^2\binom{272}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{272(272+1)\\cdot 2^{270}$.\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 272(272+1)\\cdot 2^{270}.",
"... | [
{
"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-013602 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Give an answer and a quick verification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5517} k^2\binom{5517}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,5517\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5517(5517+1)\\cdot 2^{5515}$.\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 5517(5517+1)\\cdot 2^{5515}.... | [
{
"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 $\boxed{5517(5517+1)\cdot 2^{5515}$.) |
math-013603 | Combinatorics: Binomial Sums — Double Counting | 7 | Keep the final answer in boxed form: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6238} k\binom{6238}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Bri... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,6238\\}$. 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: $\\boxed{6238\\cdot 2^{6237}$.\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-013604 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Solve and include a self-check: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3538} k^2\binom{3538}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,3538\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3538(3538+1)\\cdot 2^{3536}$.\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 3538(3538+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 $\boxed{3538(3538+1)\cdot 2^{3536}$.) |
math-013605 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Solve and then verify: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2811} k^2\binom{2811}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2811\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2811(2811+1)\\cdot 2^{2809}$.\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 2811(2811+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 $k$. |
math-013606 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Explain what is being counted/optimized: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2612} k^2\binom{2612}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2612(2612+1)\\cdot 2^{2610}$.\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 2612(2612+1)\\cdot 2^{2610}.... | [
{
"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-013607 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Track units/moduli carefully: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5615} k^2\binom{5615}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,5615\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5615(5615+1)\\cdot 2^{5613}$.\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 5615(5615+1)\\cdot 2^{5613}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{5615(5615+1)\cdot 2^{5613}$.) |
math-013608 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Use two approaches if possible: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3995} k^2\binom{3995}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3995(3995+1)\\cdot 2^{3993}$.\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 3995(3995+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 $\boxed{3995(3995+1)\cdot 2^{3993}$.) |
math-013609 | Combinatorics: Binomial Sums — Double Counting | 7 | Give an answer and a quick verification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1224} k^2\binom{1224}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1224(1224+1)\\cdot 2^{1222}$.\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 1224(1224+1)\\cdot 2^{1222}.",
"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 $\boxed{1224(1224+1)\cdot 2^{1222}$.) |
math-013610 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Explain each transformation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4881} k\binom{4881}{k}.$$
(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 $[n]=\\{1,2,\\dots,4881\\}$. 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: $\\boxed{4881\\cdot 2^{4880}$.\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-013611 | Combinatorics: Binomial Sums — Double Counting | 7 | State any required conditions first: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{790} k\binom{790}{k}.$$
(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: $\\boxed{790\\cdot 2^{789}$.\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 790\\... | [
{
"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 $\boxed{790\cdot 2^{789}$.) |
math-013612 | Combinatorics: Binomial Sums — Double Counting | 7 | Work carefully and justify each inference: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5560} k^2\binom{5560}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c)... | [
{
"method_name": "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 $A\\subseteq[n]=\\{1,\\dots,5560\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5560(5560+1)\\cdot 2^{5558}$.\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 5560(5560+1)\\cdot 2^{5558}.... | [
{
"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-013613 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Provide both a computational and a conceptual explanation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6265} k\binom{6265}{k}.$$
(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: $\\boxed{6265\\cdot 2^{6264}$.\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 626... | [
{
"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-013614 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Give a theorem-based solution: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{554} k\binom{554}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both appro... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{554\\cdot 2^{553}$.\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... | 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-013615 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Try to avoid pattern-matching; explain why: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1892} k\binom{1892}{k}.$$
(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: $\\boxed{1892\\cdot 2^{1891}$.\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 189... | [
{
"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-013616 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Give reasoning, not just computation: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1158} k\binom{1158}{k}.$$
(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 $[n]=\\{1,2,\\dots,1158\\}$. 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: $\\boxed{1158\\cdot 2^{1157}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-013617 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Exercise: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5825} k\binom{5825}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the ... | [
{
"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: $\\boxed{5825\\cdot 2^{5824}$.\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-013618 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Carefully track domains: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{261} k\binom{261}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{261\\cdot 2^{260}$.\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... | 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 $\boxed{261\cdot 2^{260}$.) |
math-013619 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Where appropriate, name the theorem you use: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6155} k\binom{6155}{k}.$$
(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 $[n]=\\{1,2,\\dots,6155\\}$. 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: $\\boxed{6155\\cdot 2^{6154}$.\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 $(1+x)^n$. |
math-013620 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Checkpoint: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4584} k^2\binom{4584}{k}.$$
(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 combin... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,4584\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4584(4584+1)\\cdot 2^{4582}$.\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 4584(4584+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 $\boxed{4584(4584+1)\cdot 2^{4582}$.) |
math-013621 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Solve (and briefly cross-validate): Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5523} k\binom{5523}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly expl... | [
{
"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 $[n]=\\{1,2,\\dots,5523\\}$. 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: $\\boxed{5523\\cdot 2^{5522}$.\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 552... | [
{
"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 $\boxed{5523\cdot 2^{5522}$.) |
math-013622 | Combinatorics: Binomial Sums — Double Counting | 7 | Answer using clear logical steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2338} k\binom{2338}{k}.$$
(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: $\\boxed{2338\\cdot 2^{2337}$.\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 $\boxed{2338\cdot 2^{2337}$.) |
math-013623 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Solve (and briefly cross-validate): Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2960} k^2\binom{2960}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c... | [
{
"method_name": "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 $A\\subseteq[n]=\\{1,\\dots,2960\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2960(2960+1)\\cdot 2^{2958}$.\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 2960(2960+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 $k$. |
math-013624 | Combinatorics: Binomial Sums — Double Counting | 7 | Solve and sanity-check: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2873} k^2\binom{2873}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully ... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2873(2873+1)\\cdot 2^{2871}$.\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 2873(2873+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-013625 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Complete the analysis: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3237} k^2\binom{3237}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{3237(3237+1)\\cdot 2^{3235}$.\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 3237(3237+1)\\cdot 2^{3235}.",
"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 $\boxed{3237(3237+1)\cdot 2^{3235}$.) |
math-013626 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Proceed methodically: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5448} k^2\binom{5448}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain care... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5448(5448+1)\\cdot 2^{5446}$.\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 5448(5448+1)\\cdot 2^{5446}.... | [
{
"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-013627 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Derive the result step-by-step: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6560} k\binom{6560}{k}.$$
(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 $[n]=\\{1,2,\\dots,6560\\}$. 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: $\\boxed{6560\\cdot 2^{6559}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-013628 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Determine the requested value: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{568} k\binom{568}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,568\\}$. 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: $\\boxed{568\\cdot 2^{567}$.\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 568\\... | [
{
"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 $\boxed{568\cdot 2^{567}$.) |
math-013629 | Combinatorics: Binomial Sums — Double Counting | 7 | Use two approaches if possible: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7282} k^2\binom{7282}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{7282(7282+1)\\cdot 2^{7280}$.\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 7282(7282+1)\\cdot 2^{7280}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-013630 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Do not skip justification steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{121} k\binom{121}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both app... | [
{
"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 $[n]=\\{1,2,\\dots,121\\}$. 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: $\\boxed{121\\cdot 2^{120}$.\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 121\\... | [
{
"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 $\boxed{121\cdot 2^{120}$.) |
math-013631 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Warm-up: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7498} k\binom{7498}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the s... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,7498\\}$. 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: $\\boxed{7498\\cdot 2^{7497}$.\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 $\boxed{7498\cdot 2^{7497}$.) |
math-013632 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Track units/moduli carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1392} k^2\binom{1392}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain careful... | [
{
"method_name": "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 $A\\subseteq[n]=\\{1,\\dots,1392\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1392(1392+1)\\cdot 2^{1390}$.\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 1392(1392+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 $\boxed{1392(1392+1)\cdot 2^{1390}$.) |
math-013633 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Exercise: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6552} k^2\binom{6552}{k}.$$
(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": "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: $\\boxed{6552(6552+1)\\cdot 2^{6550}$.\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 6552(6552+1)\\cdot 2^{6550}.... | [
{
"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 $\boxed{6552(6552+1)\cdot 2^{6550}$.) |
math-013634 | Combinatorics: Binomial Sums — Double Counting | 7 | Warm-up: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6203} k^2\binom{6203}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6203\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6203(6203+1)\\cdot 2^{6201}$.\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 6203(6203+1)\\cdot 2^{6201}.... | [
{
"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 $\boxed{6203(6203+1)\cdot 2^{6201}$.) |
math-013635 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Problem: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1223} k\binom{1223}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the s... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,1223\\}$. 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: $\\boxed{1223\\cdot 2^{1222}$.\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 122... | [
{
"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 $\boxed{1223\cdot 2^{1222}$.) |
math-013636 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Provide both a computational and a conceptual explanation: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3991} k^2\binom{3991}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family... | [
{
"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: $\\boxed{3991(3991+1)\\cdot 2^{3989}$.\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 3991(3991+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 $\boxed{3991(3991+1)\cdot 2^{3989}$.) |
math-013637 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Problem: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4593} k\binom{4593}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approa... | [
{
"method_name": "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: $\\boxed{4593\\cdot 2^{4592}$.\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 $\boxed{4593\cdot 2^{4592}$.) |
math-013638 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Prompt: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5625} k^2\binom{5625}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,5625\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5625(5625+1)\\cdot 2^{5623}$.\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 5625(5625+1)\\cdot 2^{5623}.",
"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 $\boxed{5625(5625+1)\cdot 2^{5623}$.) |
math-013639 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Compute the requested quantity: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{428} k^2\binom{428}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{428(428+1)\\cdot 2^{426}$.\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 428(428+1)\\cdot 2^{426}.",
"... | [
{
"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 $\boxed{428(428+1)\cdot 2^{426}$.) |
math-013640 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Explain each transformation: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5407} k\binom{5407}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ap... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5407\\cdot 2^{5406}$.\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 540... | [
{
"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-013641 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Solve with verification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6313} k^2\binom{6313}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6313\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6313(6313+1)\\cdot 2^{6311}$.\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 6313(6313+1)\\cdot 2^{6311}.",
"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 $\boxed{6313(6313+1)\cdot 2^{6311}$.) |
math-013642 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Challenge: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3638} k\binom{3638}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the sa... | [
{
"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 $[n]=\\{1,2,\\dots,3638\\}$. 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: $\\boxed{3638\\cdot 2^{3637}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-013643 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Derive the result step-by-step: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5935} k\binom{5935}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ap... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5935\\cdot 2^{5934}$.\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 593... | [
{
"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 $\boxed{5935\cdot 2^{5934}$.) |
math-013644 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Explain what is being counted/optimized: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6122} k^2\binom{6122}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6122\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6122(6122+1)\\cdot 2^{6120}$.\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 6122(6122+1)\\cdot 2^{6120}.",
"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-013645 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Task: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1262} k^2\binom{1262}{k}.$$
(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 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 $A\\subseteq[n]=\\{1,\\dots,1262\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1262(1262+1)\\cdot 2^{1260}$.\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 1262(1262+1)\\cdot 2^{1260}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{1262(1262+1)\cdot 2^{1260}$.) |
math-013646 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Track units/moduli carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4381} k\binom{4381}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both appr... | [
{
"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 $[n]=\\{1,2,\\dots,4381\\}$. 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: $\\boxed{4381\\cdot 2^{4380}$.\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 438... | [
{
"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 $\boxed{4381\cdot 2^{4380}$.) |
math-013647 | Combinatorics: Binomial Sums — Double Counting | 7 | Complete the analysis: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4087} k^2\binom{4087}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Exp... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4087(4087+1)\\cdot 2^{4085}$.\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 4087(4087+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 $\boxed{4087(4087+1)\cdot 2^{4085}$.) |
math-013648 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Make each step logically reversible (or explain if not): Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4649} k\binom{4649}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argum... | [
{
"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: $\\boxed{4649\\cdot 2^{4648}$.\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 $\boxed{4649\cdot 2^{4648}$.) |
math-013649 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Find the exact value: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3335} k^2\binom{3335}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,3335\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3335(3335+1)\\cdot 2^{3333}$.\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 3335(3335+1)\\cdot 2^{3333}.",
"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-013650 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Explain each transformation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6483} k^2\binom{6483}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{6483(6483+1)\\cdot 2^{6481}$.\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 6483(6483+1)\\cdot 2^{6481}.",
"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 $\boxed{6483(6483+1)\cdot 2^{6481}$.) |
math-013651 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Start by stating any domain restrictions: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3558} k^2\binom{3558}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Exp... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3558(3558+1)\\cdot 2^{3556}$.\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 3558(3558+1)\\cdot 2^{3556}.... | [
{
"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 $\boxed{3558(3558+1)\cdot 2^{3556}$.) |
math-013652 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Give a fully justified solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1037} k^2\binom{1037}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{1037(1037+1)\\cdot 2^{1035}$.\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 1037(1037+1)\\cdot 2^{1035}.",
"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-013653 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Solve and sanity-check: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7570} k\binom{7570}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7570\\cdot 2^{7569}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-013654 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Work carefully and justify each inference: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2528} k\binom{2528}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly expla... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,2528\\}$. 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: $\\boxed{2528\\cdot 2^{2527}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{2528\cdot 2^{2527}$.) |
math-013655 | Combinatorics: Binomial Sums — Double Counting | 7 | Solve and sanity-check: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{174} k^2\binom{174}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,174\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{174(174+1)\\cdot 2^{172}$.\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 174(174+1)\\cdot 2^{172}.",
"... | [
{
"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 $\boxed{174(174+1)\cdot 2^{172}$.) |
math-013656 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Explain each transformation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3929} k\binom{3929}{k}.$$
(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 $[n]=\\{1,2,\\dots,3929\\}$. 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: $\\boxed{3929\\cdot 2^{3928}$.\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-013657 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Warm-up: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3336} k\binom{3336}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the s... | [
{
"method_name": "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: $\\boxed{3336\\cdot 2^{3335}$.\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 $\boxed{3336\cdot 2^{3335}$.) |
math-013658 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Try to avoid pattern-matching; explain why: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2936} k\binom{2936}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2936\\cdot 2^{2935}$.\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-013659 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Carefully track domains: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6532} k\binom{6532}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approache... | [
{
"method_name": "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: $\\boxed{6532\\cdot 2^{6531}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-013660 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Give a theorem-based solution: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3033} k^2\binom{3033}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,3033\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3033(3033+1)\\cdot 2^{3031}$.\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 3033(3033+1)\\cdot 2^{3031}.... | [
{
"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 $\boxed{3033(3033+1)\cdot 2^{3031}$.) |
math-013661 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Work carefully and justify each inference: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5739} k^2\binom{5739}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,5739\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5739(5739+1)\\cdot 2^{5737}$.\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 5739(5739+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 $\boxed{5739(5739+1)\cdot 2^{5737}$.) |
math-013662 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Use two approaches if possible: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1909} k^2\binom{1909}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,1909\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1909(1909+1)\\cdot 2^{1907}$.\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 1909(1909+1)\\cdot 2^{1907}.... | [
{
"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-013663 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Task: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5440} k\binom{5440}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approache... | [
{
"method_name": "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: $\\boxed{5440\\cdot 2^{5439}$.\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 $\boxed{5440\cdot 2^{5439}$.) |
math-013664 | Combinatorics: Binomial Sums — Double Counting | 7 | Explain each transformation: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5034} k\binom{5034}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,5034\\}$. 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: $\\boxed{5034\\cdot 2^{5033}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 503... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-013665 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Provide a rigorous solution: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2559} k^2\binom{2559}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{2559(2559+1)\\cdot 2^{2557}$.\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 2559(2559+1)\\cdot 2^{2557}.",
"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 $\boxed{2559(2559+1)\cdot 2^{2557}$.) |
math-013666 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Provide both a computational and a conceptual explanation: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3497} k^2\binom{3497}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,3497\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3497(3497+1)\\cdot 2^{3495}$.\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 3497(3497+1)\\cdot 2^{3495}.",
"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 $\boxed{3497(3497+1)\cdot 2^{3495}$.) |
math-013667 | Combinatorics: Binomial Sums — Double Counting | 7 | Give a fully justified solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5959} k\binom{5959}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5959\\cdot 2^{5958}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{5959\cdot 2^{5958}$.) |
math-013668 | Combinatorics: Binomial Sums — Double Counting | 7 | Solve (and briefly cross-validate): Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7445} k^2\binom{7445}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{7445(7445+1)\\cdot 2^{7443}$.\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 7445(7445+1)\\cdot 2^{7443}.",
"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 $\boxed{7445(7445+1)\cdot 2^{7443}$.) |
math-013669 | Combinatorics: Binomial Sums — Double Counting | 7 | Provide both a computational and a conceptual explanation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7150} k^2\binom{7150}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an ap... | [
{
"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: $\\boxed{7150(7150+1)\\cdot 2^{7148}$.\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 7150(7150+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-013670 | Combinatorics: Binomial Sums — Double Counting | 7 | Derive the result step-by-step: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3199} k\binom{3199}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain ... | [
{
"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 $[n]=\\{1,2,\\dots,3199\\}$. 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: $\\boxed{3199\\cdot 2^{3198}$.\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 $\boxed{3199\cdot 2^{3198}$.) |
math-013671 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Prompt: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7194} k\binom{7194}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approac... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7194\\cdot 2^{7193}$.\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 $\boxed{7194\cdot 2^{7193}$.) |
math-013672 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Compute the requested quantity: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7273} k^2\binom{7273}{k}.$$
(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: $\\boxed{7273(7273+1)\\cdot 2^{7271}$.\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 7273(7273+1)\\cdot 2^{7271}.",
"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 $\boxed{7273(7273+1)\cdot 2^{7271}$.) |
math-013673 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Derive the result step-by-step: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2483} k\binom{2483}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2483\\cdot 2^{2482}$.\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 $\boxed{2483\cdot 2^{2482}$.) |
math-013674 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Complete the analysis: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1107} k\binom{1107}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approach... | [
{
"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: $\\boxed{1107\\cdot 2^{1106}$.\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 $\boxed{1107\cdot 2^{1106}$.) |
math-013675 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Indicate where a theorem is used: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2658} k^2\binom{2658}{k}.$$
(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: $\\boxed{2658(2658+1)\\cdot 2^{2656}$.\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 2658(2658+1)\\cdot 2^{2656}.",
"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 $\boxed{2658(2658+1)\cdot 2^{2656}$.) |
math-013676 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Be explicit about assumptions: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6829} k\binom{6829}{k}.$$
(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 $[n]=\\{1,2,\\dots,6829\\}$. 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: $\\boxed{6829\\cdot 2^{6828}$.\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 $\boxed{6829\cdot 2^{6828}$.) |
math-013677 | Combinatorics: Binomial Sums — Double Counting | 7 | Challenge: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7359} k^2\binom{7359}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,7359\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7359(7359+1)\\cdot 2^{7357}$.\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 7359(7359+1)\\cdot 2^{7357}.... | [
{
"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 $\boxed{7359(7359+1)\cdot 2^{7357}$.) |
math-013678 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Find the exact value: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3599} k^2\binom{3599}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3599(3599+1)\\cdot 2^{3597}$.\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 3599(3599+1)\\cdot 2^{3597}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-013679 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | State any required conditions first: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5437} k\binom{5437}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{5437\\cdot 2^{5436}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{5437\cdot 2^{5436}$.) |
math-013680 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Complete the analysis: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2037} k^2\binom{2037}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2037\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2037(2037+1)\\cdot 2^{2035}$.\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 2037(2037+1)\\cdot 2^{2035}.",
"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 $\boxed{2037(2037+1)\cdot 2^{2035}$.) |
math-013681 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Solve and justify each step: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5599} k\binom{5599}{k}.$$
(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 $[n]=\\{1,2,\\dots,5599\\}$. 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: $\\boxed{5599\\cdot 2^{5598}$.\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-013682 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Start by stating any domain restrictions: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{667} k^2\binom{667}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,667\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{667(667+1)\\cdot 2^{665}$.\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 667(667+1)\\cdot 2^{66... | [
{
"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-013683 | Combinatorics: Binomial Sums — Double Counting | 7 | 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
$$S=\sum_{k=0}^{7178} k\binom{7178}{k}.$$
(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: $\\boxed{7178\\cdot 2^{7177}$.\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-013684 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Write the solution set clearly: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1589} k\binom{1589}{k}.$$
(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: $\\boxed{1589\\cdot 2^{1588}$.\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 158... | [
{
"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-013685 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Determine the requested value: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1171} k\binom{1171}{k}.$$
(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 $[n]=\\{1,2,\\dots,1171\\}$. 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: $\\boxed{1171\\cdot 2^{1170}$.\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 $\boxed{1171\cdot 2^{1170}$.) |
math-013686 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Exercise: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3041} k\binom{3041}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{3041\\cdot 2^{3040}$.\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 304... | [
{
"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-013687 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Provide both a computational and a conceptual explanation: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4730} k^2\binom{4730}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriat... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,4730\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4730(4730+1)\\cdot 2^{4728}$.\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 4730(4730+1)\\cdot 2^{4728}.",
"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 $\boxed{4730(4730+1)\cdot 2^{4728}$.) |
math-013688 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Write the solution set clearly: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7461} k\binom{7461}{k}.$$
(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: $\\boxed{7461\\cdot 2^{7460}$.\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 746... | [
{
"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 $\boxed{7461\cdot 2^{7460}$.) |
math-013689 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Provide a rigorous solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5200} k\binom{5200}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,5200\\}$. 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: $\\boxed{5200\\cdot 2^{5199}$.\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-013690 | Combinatorics: Binomial Sums — Double Counting | 7 | Task: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6915} k\binom{6915}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the same qu... | [
{
"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 $[n]=\\{1,2,\\dots,6915\\}$. 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: $\\boxed{6915\\cdot 2^{6914}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-013691 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Task: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6584} k^2\binom{6584}{k}.$$
(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 combinatoria... | [
{
"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: $\\boxed{6584(6584+1)\\cdot 2^{6582}$.\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 6584(6584+1)\\cdot 2^{6582}.",
"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 $\boxed{6584(6584+1)\cdot 2^{6582}$.) |
math-013692 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Do not skip justification steps: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2963} k\binom{2963}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly... | [
{
"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 $[n]=\\{1,2,\\dots,2963\\}$. 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: $\\boxed{2963\\cdot 2^{2962}$.\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-013693 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 7 | Do not skip justification steps: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2029} k^2\binom{2029}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2029\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2029(2029+1)\\cdot 2^{2027}$.\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 2029(2029+1)\\cdot 2^{2027}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-013694 | Combinatorics: Binomial Sums — Double Counting | 7 | State any required conditions first: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1571} k^2\binom{1571}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,1571\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1571(1571+1)\\cdot 2^{1569}$.\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 1571(1571+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 $\boxed{1571(1571+1)\cdot 2^{1569}$.) |
math-013695 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Answer with a short justification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1848} k^2\binom{1848}{k}.$$
(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: $\\boxed{1848(1848+1)\\cdot 2^{1846}$.\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 1848(1848+1)\\cdot 2^{1846}.",
"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 $\boxed{1848(1848+1)\cdot 2^{1846}$.) |
math-013696 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 7 | Solve and justify each step: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4118} k^2\binom{4118}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefull... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4118(4118+1)\\cdot 2^{4116}$.\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 4118(4118+1)\\cdot 2^{4116}.... | [
{
"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 $\boxed{4118(4118+1)\cdot 2^{4116}$.) |
math-013697 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Exercise: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1348} k\binom{1348}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the ... | [
{
"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 $[n]=\\{1,2,\\dots,1348\\}$. 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: $\\boxed{1348\\cdot 2^{1347}$.\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 $(1+x)^n$. (Here the result is $\boxed{1348\cdot 2^{1347}$.) |
math-013698 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 7 | Do not skip justification steps: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1231} k^2\binom{1231}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,1231\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1231(1231+1)\\cdot 2^{1229}$.\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 1231(1231+1)\\cdot 2^{1229}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Remember: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-013699 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Derive the result step-by-step: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{185} k^2\binom{185}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,185\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{185(185+1)\\cdot 2^{183}$.\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 185(185+1)\\cdot 2^{183}.",
"... | [
{
"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 $\boxed{185(185+1)\cdot 2^{183}$.) |
math-013700 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 7 | Answer with a short justification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1239} k^2\binom{1239}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain ca... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1239(1239+1)\\cdot 2^{1237}$.\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 1239(1239+1)\\cdot 2^{1237}.",
"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 $\boxed{1239(1239+1)\cdot 2^{1237}$.) |
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