id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-019501 | Combinatorics: Binomial Sums — Double Counting | 10 | Where appropriate, name the theorem you use: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7862} k\binom{7862}{k}.$$
(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 $[n]=\\{1,2,\\dots,7862\\}$. 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: $\\boxed{7862\\cdot 2^{7861}$.\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 $\boxed{7862\cdot 2^{7861}$.) |
math-019502 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Challenge: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{95} k^2\binom{95}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully w... | [
{
"method_name": "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: $\\boxed{95(95+1)\\cdot 2^{93}$.\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 95(95+1)\\cdot 2^{93}.",
... | [
{
"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 $\boxed{95(95+1)\cdot 2^{93}$.) |
math-019503 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Problem: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5623} k^2\binom{5623}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5623(5623+1)\\cdot 2^{5621}$.\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 5623(5623+1)\\cdot 2^{5621}.",
"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 $\boxed{5623(5623+1)\cdot 2^{5621}$.) |
math-019504 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Give a theorem-based solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7085} k^2\binom{7085}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain car... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{7085(7085+1)\\cdot 2^{7083}$.\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 7085(7085+1)\\cdot 2^{7083}.",
"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 $\boxed{7085(7085+1)\cdot 2^{7083}$.) |
math-019505 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Determine the requested value: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1906} k^2\binom{1906}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain car... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1906(1906+1)\\cdot 2^{1904}$.\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 1906(1906+1)\\cdot 2^{1904}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-019506 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Use two approaches if possible: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1740} k^2\binom{1740}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1740(1740+1)\\cdot 2^{1738}$.\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 1740(1740+1)\\cdot 2^{1738}.... | [
{
"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 $\boxed{1740(1740+1)\cdot 2^{1738}$.) |
math-019507 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Answer with a short justification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7321} k^2\binom{7321}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain... | [
{
"method_name": "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: $\\boxed{7321(7321+1)\\cdot 2^{7319}$.\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 7321(7321+1)\\cdot 2^{7319}.... | [
{
"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 $\boxed{7321(7321+1)\cdot 2^{7319}$.) |
math-019508 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Find the exact value: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5488} k^2\binom{5488}{k}.$$
(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: $\\boxed{5488(5488+1)\\cdot 2^{5486}$.\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 5488(5488+1)\\cdot 2^{5486}.",
"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-019509 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Make each step logically reversible (or explain if not): Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2747} k\binom{2747}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counti... | [
{
"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: $\\boxed{2747\\cdot 2^{2746}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-019510 | Combinatorics: Binomial Sums — Double Counting | 10 | Work this out carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6645} k\binom{6645}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approache... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,6645\\}$. 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: $\\boxed{6645\\cdot 2^{6644}$.\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 $(1+x)^n$. |
math-019511 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Explain each transformation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2407} k^2\binom{2407}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefull... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2407\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2407(2407+1)\\cdot 2^{2405}$.\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 2407(2407+1)\\cdot 2^{2405}.... | [
{
"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 $\boxed{2407(2407+1)\cdot 2^{2405}$.) |
math-019512 | Combinatorics: Binomial Sums — Double Counting | 10 | Complete the analysis: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7102} k\binom{7102}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches ... | [
{
"method_name": "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 $[n]=\\{1,2,\\dots,7102\\}$. 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: $\\boxed{7102\\cdot 2^{7101}$.\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 710... | [
{
"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 $\boxed{7102\cdot 2^{7101}$.) |
math-019513 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Be explicit about assumptions: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4492} k\binom{4492}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4492\\cdot 2^{4491}$.\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-019514 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | State any required conditions first: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3657} k\binom{3657}{k}.$$
(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 $[n]=\\{1,2,\\dots,3657\\}$. 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: $\\boxed{3657\\cdot 2^{3656}$.\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 $\boxed{3657\cdot 2^{3656}$.) |
math-019515 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Exercise: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5852} k\binom{5852}{k}.$$
(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": "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 $[n]=\\{1,2,\\dots,5852\\}$. 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: $\\boxed{5852\\cdot 2^{5851}$.\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 585... | [
{
"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 $\boxed{5852\cdot 2^{5851}$.) |
math-019516 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Start by stating any domain restrictions: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5384} k\binom{5384}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,5384\\}$. 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: $\\boxed{5384\\cdot 2^{5383}$.\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 538... | [
{
"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 $\boxed{5384\cdot 2^{5383}$.) |
math-019517 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | State any required conditions first: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6373} k\binom{6373}{k}.$$
(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 $[n]=\\{1,2,\\dots,6373\\}$. 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: $\\boxed{6373\\cdot 2^{6372}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-019518 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Compute the requested quantity: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5267} k^2\binom{5267}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,5267\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5267(5267+1)\\cdot 2^{5265}$.\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 5267(5267+1)\\cdot 2^{5265}.",
"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 $\boxed{5267(5267+1)\cdot 2^{5265}$.) |
math-019519 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Give a theorem-based solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1560} k^2\binom{1560}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain car... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1560(1560+1)\\cdot 2^{1558}$.\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 1560(1560+1)\\cdot 2^{1558}.",
"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-019520 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Work carefully and justify each inference: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2215} k^2\binom{2215}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of trip... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2215\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2215(2215+1)\\cdot 2^{2213}$.\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 2215(2215+1)\\cdot 2^{2213}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-019521 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Work this out carefully: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3979} k^2\binom{3979}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain c... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3979(3979+1)\\cdot 2^{3977}$.\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 3979(3979+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-019522 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Complete the analysis: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7020} k\binom{7020}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7020\\cdot 2^{7019}$.\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 $\boxed{7020\cdot 2^{7019}$.) |
math-019523 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Give a fully justified solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{537} k^2\binom{537}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain car... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{537(537+1)\\cdot 2^{535}$.\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 537(537+1)\\cdot 2^{535}.",
"robustness_analysis": "... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | 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 $\boxed{537(537+1)\cdot 2^{535}$.) |
math-019524 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Track units/moduli carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7527} k^2\binom{7527}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain careful... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7527(7527+1)\\cdot 2^{7525}$.\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 7527(7527+1)\\cdot 2^{7525}.",
"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-019525 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Provide both a computational and a conceptual explanation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5210} k\binom{5210}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) ... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,5210\\}$. 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: $\\boxed{5210\\cdot 2^{5209}$.\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 521... | [
{
"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 $\boxed{5210\cdot 2^{5209}$.) |
math-019526 | Combinatorics: Binomial Sums — Double Counting | 10 | Work this out carefully: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2165} k^2\binom{2165}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2165\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2165(2165+1)\\cdot 2^{2163}$.\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 2165(2165+1)\\cdot 2^{2163}.... | [
{
"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 $\boxed{2165(2165+1)\cdot 2^{2163}$.) |
math-019527 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Solve and sanity-check: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5425} k^2\binom{5425}{k}.$$
(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: $\\boxed{5425(5425+1)\\cdot 2^{5423}$.\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 5425(5425+1)\\cdot 2^{5423}.",
"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-019528 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Problem: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7529} k^2\binom{7529}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why your combinato... | [
{
"method_name": "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: $\\boxed{7529(7529+1)\\cdot 2^{7527}$.\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 7529(7529+1)\\cdot 2^{7527}.",
"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 $\boxed{7529(7529+1)\cdot 2^{7527}$.) |
math-019529 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Explain what is being counted/optimized: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2721} k\binom{2721}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c)... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2721\\cdot 2^{2720}$.\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 $\boxed{2721\cdot 2^{2720}$.) |
math-019530 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Problem: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6359} k^2\binom{6359}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6359\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6359(6359+1)\\cdot 2^{6357}$.\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 6359(6359+1)\\cdot 2^{6357}.",
"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 $\boxed{6359(6359+1)\cdot 2^{6357}$.) |
math-019531 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Try to avoid pattern-matching; explain why: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3771} k\binom{3771}{k}.$$
(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 $[n]=\\{1,2,\\dots,3771\\}$. 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: $\\boxed{3771\\cdot 2^{3770}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-019532 | Combinatorics: Binomial Sums — Double Counting | 10 | Solve and include a self-check: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{2517} k\binom{2517}{k}.$$
(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 $[n]=\\{1,2,\\dots,2517\\}$. 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: $\\boxed{2517\\cdot 2^{2516}$.\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 $\boxed{2517\cdot 2^{2516}$.) |
math-019533 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | State any required conditions first: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4113} k^2\binom{4113}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tr... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4113\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4113(4113+1)\\cdot 2^{4111}$.\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 4113(4113+1)\\cdot 2^{4111}.",
"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 $\boxed{4113(4113+1)\cdot 2^{4111}$.) |
math-019534 | Combinatorics: Binomial Sums — Double Counting | 10 | Give reasoning, not just computation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5323} k\binom{5323}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Br... | [
{
"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 $[n]=\\{1,2,\\dots,5323\\}$. 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: $\\boxed{5323\\cdot 2^{5322}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{5323\cdot 2^{5322}$.) |
math-019535 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Do not skip justification steps: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5150} k^2\binom{5150}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triple... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,5150\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5150(5150+1)\\cdot 2^{5148}$.\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 5150(5150+1)\\cdot 2^{5148}.... | [
{
"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 $\boxed{5150(5150+1)\cdot 2^{5148}$.) |
math-019536 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Explain why your operations are valid: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6835} k^2\binom{6835}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Exp... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,6835\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6835(6835+1)\\cdot 2^{6833}$.\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 6835(6835+1)\\cdot 2^{6833}.",
"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 $\boxed{6835(6835+1)\cdot 2^{6833}$.) |
math-019537 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Problem: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7224} k\binom{7224}{k}.$$
(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... | [
{
"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: $\\boxed{7224\\cdot 2^{7223}$.\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-019538 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Question: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1719} k^2\binom{1719}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{1719(1719+1)\\cdot 2^{1717}$.\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 1719(1719+1)\\cdot 2^{1717}.",
"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 $\boxed{1719(1719+1)\cdot 2^{1717}$.) |
math-019539 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Checkpoint: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{512} k\binom{512}{k}.$$
(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": "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 $[n]=\\{1,2,\\dots,512\\}$. 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: $\\boxed{512\\cdot 2^{511}$.\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 512\\... | [
{
"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 $\boxed{512\cdot 2^{511}$.) |
math-019540 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Track units/moduli carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4172} k\binom{4172}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both a... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4172\\cdot 2^{4171}$.\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 $\boxed{4172\cdot 2^{4171}$.) |
math-019541 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Compute the requested quantity: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5552} k^2\binom{5552}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,5552\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5552(5552+1)\\cdot 2^{5550}$.\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 5552(5552+1)\\cdot 2^{5550}.... | [
{
"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 $\boxed{5552(5552+1)\cdot 2^{5550}$.) |
math-019542 | Combinatorics: Binomial Sums — Double Counting | 10 | Track quantifiers carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4947} k^2\binom{4947}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefull... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4947\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4947(4947+1)\\cdot 2^{4945}$.\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 4947(4947+1)\\cdot 2^{4945}.... | [
{
"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 $\boxed{4947(4947+1)\cdot 2^{4945}$.) |
math-019543 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Keep the final answer in boxed form: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2717} k\binom{2717}{k}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{2717\\cdot 2^{2716}$.\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 271... | [
{
"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-019544 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Solve and sanity-check: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2957} k^2\binom{2957}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,2957\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2957(2957+1)\\cdot 2^{2955}$.\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 2957(2957+1)\\cdot 2^{2955}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-019545 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Give a fully justified solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6819} k^2\binom{6819}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6819\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6819(6819+1)\\cdot 2^{6817}$.\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 6819(6819+1)\\cdot 2^{6817}.",
"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 $\boxed{6819(6819+1)\cdot 2^{6817}$.) |
math-019546 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Explain each transformation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1890} k\binom{1890}{k}.$$
(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 $[n]=\\{1,2,\\dots,1890\\}$. 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: $\\boxed{1890\\cdot 2^{1889}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-019547 | Combinatorics: Binomial Sums — Double Counting | 10 | Give an answer and a quick verification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{909} k^2\binom{909}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of ... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{909(909+1)\\cdot 2^{907}$.\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 909(909+1)\\cdot 2^{90... | [
{
"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 $\boxed{909(909+1)\cdot 2^{907}$.) |
math-019548 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Give an answer and a quick verification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1015} k^2\binom{1015}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triple... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1015\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1015(1015+1)\\cdot 2^{1013}$.\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 1015(1015+1)\\cdot 2^{1013}.",
"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 $\boxed{1015(1015+1)\cdot 2^{1013}$.) |
math-019549 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Explain each transformation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7887} k^2\binom{7887}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefull... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7887(7887+1)\\cdot 2^{7885}$.\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 7887(7887+1)\\cdot 2^{7885}.",
"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 $\boxed{7887(7887+1)\cdot 2^{7885}$.) |
math-019550 | Combinatorics: Binomial Sums — Double Counting | 10 | Keep the final answer in boxed form: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7465} k^2\binom{7465}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain ... | [
{
"method_name": "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: $\\boxed{7465(7465+1)\\cdot 2^{7463}$.\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 7465(7465+1)\\cdot 2^{7463}.",
"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 $\boxed{7465(7465+1)\cdot 2^{7463}$.) |
math-019551 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Give a fully justified solution: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{646} k^2\binom{646}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,646\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{646(646+1)\\cdot 2^{644}$.\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 646(646+1)\\cdot 2^{644}.",
"... | [
{
"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 $\boxed{646(646+1)\cdot 2^{644}$.) |
math-019552 | Combinatorics: Binomial Sums — Double Counting | 10 | Write the solution set clearly: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{113} k\binom{113}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both a... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,113\\}$. 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: $\\boxed{113\\cdot 2^{112}$.\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 113\\... | [
{
"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 $\boxed{113\cdot 2^{112}$.) |
math-019553 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | State any required conditions first: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6780} k\binom{6780}{k}.$$
(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 $[n]=\\{1,2,\\dots,6780\\}$. 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: $\\boxed{6780\\cdot 2^{6779}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-019554 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Track quantifiers carefully: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2560} k^2\binom{2560}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2560(2560+1)\\cdot 2^{2558}$.\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 2560(2560+1)\\cdot 2^{2558}.",
"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 $\boxed{2560(2560+1)\cdot 2^{2558}$.) |
math-019555 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Give a fully justified solution: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5920} k^2\binom{5920}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain c... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5920(5920+1)\\cdot 2^{5918}$.\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 5920(5920+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 $\boxed{5920(5920+1)\cdot 2^{5918}$.) |
math-019556 | Combinatorics: Binomial Sums — Double Counting | 10 | Prompt: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2905} k^2\binom{2905}{k}.$$
(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... | [
{
"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: $\\boxed{2905(2905+1)\\cdot 2^{2903}$.\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 2905(2905+1)\\cdot 2^{2903}.",
"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-019557 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Warm-up: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6067} k\binom{6067}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches cou... | [
{
"method_name": "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 $[n]=\\{1,2,\\dots,6067\\}$. 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: $\\boxed{6067\\cdot 2^{6066}$.\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-019558 | Combinatorics: Binomial Sums — Double Counting | 10 | Proceed methodically: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5706} k\binom{5706}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches c... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5706\\cdot 2^{5705}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-019559 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Work carefully and justify each inference: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2191} k^2\binom{2191}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of trip... | [
{
"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: $\\boxed{2191(2191+1)\\cdot 2^{2189}$.\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 2191(2191+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{2191(2191+1)\cdot 2^{2189}$.) |
math-019560 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Solve and justify each step: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{2463} k^2\binom{2463}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expla... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2463(2463+1)\\cdot 2^{2461}$.\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 2463(2463+1)\\cdot 2^{2461}.... | [
{
"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-019561 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Keep the final answer in boxed form: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5224} k^2\binom{5224}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tr... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5224(5224+1)\\cdot 2^{5222}$.\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 5224(5224+1)\\cdot 2^{5222}.",
"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-019562 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Prompt: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1951} k\binom{1951}{k}.$$
(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 ... | [
{
"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 $[n]=\\{1,2,\\dots,1951\\}$. 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: $\\boxed{1951\\cdot 2^{1950}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{1951\cdot 2^{1950}$.) |
math-019563 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Work carefully and justify each inference: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6666} k\binom{6666}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Brief... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6666\\cdot 2^{6665}$.\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 666... | [
{
"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 $\boxed{6666\cdot 2^{6665}$.) |
math-019564 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Challenge: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7204} k^2\binom{7204}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why your comb... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,7204\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7204(7204+1)\\cdot 2^{7202}$.\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 7204(7204+1)\\cdot 2^{7202}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-019565 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Work carefully and justify each inference: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5418} k\binom{5418}{k}.$$
(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 $[n]=\\{1,2,\\dots,5418\\}$. 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: $\\boxed{5418\\cdot 2^{5417}$.\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 $\boxed{5418\cdot 2^{5417}$.) |
math-019566 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Try to avoid pattern-matching; explain why: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3968} k^2\binom{3968}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,3968\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3968(3968+1)\\cdot 2^{3966}$.\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 3968(3968+1)\\cdot 2^{3966}.... | [
{
"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 $\boxed{3968(3968+1)\cdot 2^{3966}$.) |
math-019567 | Combinatorics: Binomial Sums — Double Counting | 10 | Proceed methodically: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3379} k\binom{3379}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approache... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,3379\\}$. 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: $\\boxed{3379\\cdot 2^{3378}$.\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 337... | [
{
"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 $\boxed{3379\cdot 2^{3378}$.) |
math-019568 | Combinatorics: Binomial Sums — Double Counting | 10 | Give reasoning, not just computation: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5735} k^2\binom{5735}{k}.$$
(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: $\\boxed{5735(5735+1)\\cdot 2^{5733}$.\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 5735(5735+1)\\cdot 2^{5733}.... | [
{
"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 $\boxed{5735(5735+1)\cdot 2^{5733}$.) |
math-019569 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Checkpoint: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4925} k^2\binom{4925}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why your com... | [
{
"method_name": "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 $A\\subseteq[n]=\\{1,\\dots,4925\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4925(4925+1)\\cdot 2^{4923}$.\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 4925(4925+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-019570 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Derive the result step-by-step: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3175} k\binom{3175}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3175\\cdot 2^{3174}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{3175\cdot 2^{3174}$.) |
math-019571 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Write the solution set clearly: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5618} k^2\binom{5618}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,5618\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5618(5618+1)\\cdot 2^{5616}$.\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 5618(5618+1)\\cdot 2^{5616}.",
"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 $\boxed{5618(5618+1)\cdot 2^{5616}$.) |
math-019572 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Complete the analysis: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3834} k^2\binom{3834}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,3834\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3834(3834+1)\\cdot 2^{3832}$.\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 3834(3834+1)\\cdot 2^{3832}.... | [
{
"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 $\boxed{3834(3834+1)\cdot 2^{3832}$.) |
math-019573 | Combinatorics: Binomial Sums — Double Counting | 10 | Compute the requested quantity: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7562} k\binom{7562}{k}.$$
(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 $[n]=\\{1,2,\\dots,7562\\}$. 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: $\\boxed{7562\\cdot 2^{7561}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-019574 | Combinatorics: Binomial Sums — Double Counting | 10 | Work this out carefully: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6977} k^2\binom{6977}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain c... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6977(6977+1)\\cdot 2^{6975}$.\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 6977(6977+1)\\cdot 2^{6975}.",
"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 $\boxed{6977(6977+1)\cdot 2^{6975}$.) |
math-019575 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Task: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{102} k^2\binom{102}{k}.$$
(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 combinatori... | [
{
"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 $A\\subseteq[n]=\\{1,\\dots,102\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{102(102+1)\\cdot 2^{100}$.\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 102(102+1)\\cdot 2^{100}.",
"robustness_analysis": "... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Key idea: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{102(102+1)\cdot 2^{100}$.) |
math-019576 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Task: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6698} k\binom{6698}{k}.$$
(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... | [
{
"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: $\\boxed{6698\\cdot 2^{6697}$.\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 $(1+x)^n$. |
math-019577 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Solve and justify each step: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4590} k\binom{4590}{k}.$$
(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 $[n]=\\{1,2,\\dots,4590\\}$. 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: $\\boxed{4590\\cdot 2^{4589}$.\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 459... | [
{
"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 $\boxed{4590\cdot 2^{4589}$.) |
math-019578 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Make each step logically reversible (or explain if not): Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7712} k^2\binom{7712}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of t... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{7712(7712+1)\\cdot 2^{7710}$.\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 7712(7712+1)\\cdot 2^{7710}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-019579 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Task: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5325} k\binom{5325}{k}.$$
(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": "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: $\\boxed{5325\\cdot 2^{5324}$.\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 532... | [
{
"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 $\boxed{5325\cdot 2^{5324}$.) |
math-019580 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Do not skip justification steps: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6752} k\binom{6752}{k}.$$
(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 $[n]=\\{1,2,\\dots,6752\\}$. 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: $\\boxed{6752\\cdot 2^{6751}$.\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-019581 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Explain each transformation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1926} k^2\binom{1926}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1926(1926+1)\\cdot 2^{1924}$.\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 1926(1926+1)\\cdot 2^{1924}.",
"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-019582 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Derive the result step-by-step: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2010} k\binom{2010}{k}.$$
(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: $\\boxed{2010\\cdot 2^{2009}$.\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 201... | [
{
"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-019583 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | 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
$$S=\sum_{k=0}^{2043} k^2\binom{2043}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tr... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2043(2043+1)\\cdot 2^{2041}$.\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 2043(2043+1)\\cdot 2^{2041}.",
"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-019584 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Explain what is being counted/optimized: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3047} k\binom{3047}{k}.$$
(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 $[n]=\\{1,2,\\dots,3047\\}$. 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: $\\boxed{3047\\cdot 2^{3046}$.\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 $(1+x)^n$. |
math-019585 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Start by stating any domain restrictions: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6719} k^2\binom{6719}{k}.$$
(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: $\\boxed{6719(6719+1)\\cdot 2^{6717}$.\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 6719(6719+1)\\cdot 2^{6717}.... | [
{
"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-019586 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Answer with a short justification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4710} k^2\binom{4710}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4710\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4710(4710+1)\\cdot 2^{4708}$.\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 4710(4710+1)\\cdot 2^{4708}.",
"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-019587 | Combinatorics: Binomial Sums — Double Counting | 10 | Warm-up: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1850} k\binom{1850}{k}.$$
(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 $[n]=\\{1,2,\\dots,1850\\}$. 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: $\\boxed{1850\\cdot 2^{1849}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-019588 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Answer using clear logical steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4517} k\binom{4517}{k}.$$
(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 $[n]=\\{1,2,\\dots,4517\\}$. 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: $\\boxed{4517\\cdot 2^{4516}$.\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 $\boxed{4517\cdot 2^{4516}$.) |
math-019589 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Indicate where a theorem is used: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1255} k\binom{1255}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefl... | [
{
"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 $[n]=\\{1,2,\\dots,1255\\}$. 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: $\\boxed{1255\\cdot 2^{1254}$.\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 $\boxed{1255\cdot 2^{1254}$.) |
math-019590 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Provide a rigorous solution: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5965} k\binom{5965}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly exp... | [
{
"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: $\\boxed{5965\\cdot 2^{5964}$.\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 596... | [
{
"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-019591 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Do not skip justification steps: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1873} k^2\binom{1873}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triple... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1873(1873+1)\\cdot 2^{1871}$.\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 1873(1873+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 $\boxed{1873(1873+1)\cdot 2^{1871}$.) |
math-019592 | Combinatorics: Binomial Sums — Double Counting | 10 | Answer with a short justification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{968} k^2\binom{968}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain c... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{968(968+1)\\cdot 2^{966}$.\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 968(968+1)\\cdot 2^{966}.",
"... | [
{
"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-019593 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Compute the requested quantity: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5706} k^2\binom{5706}{k}.$$
(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: $\\boxed{5706(5706+1)\\cdot 2^{5704}$.\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 5706(5706+1)\\cdot 2^{5704}.",
"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 $\boxed{5706(5706+1)\cdot 2^{5704}$.) |
math-019594 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Solve (and briefly cross-validate): Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{885} k^2\binom{885}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,885\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{885(885+1)\\cdot 2^{883}$.\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 885(885+1)\\cdot 2^{883}.",
"robustness_analysis": "... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | 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-019595 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 10 | Use two approaches if possible: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5907} k^2\binom{5907}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,5907\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5907(5907+1)\\cdot 2^{5905}$.\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 5907(5907+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-019596 | Combinatorics: Binomial Sums — Double Counting | 10 | Find the exact value: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2183} k\binom{2183}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain wh... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2183\\cdot 2^{2182}$.\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 $\boxed{2183\cdot 2^{2182}$.) |
math-019597 | Combinatorics: Binomial Sums — Double Counting | 10 | Answer using clear logical steps: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6562} k^2\binom{6562}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6562\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6562(6562+1)\\cdot 2^{6560}$.\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 6562(6562+1)\\cdot 2^{6560}.",
"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 $\boxed{6562(6562+1)\cdot 2^{6560}$.) |
math-019598 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 10 | Track units/moduli carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7271} k\binom{7271}{k}.$$
(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 $[n]=\\{1,2,\\dots,7271\\}$. 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: $\\boxed{7271\\cdot 2^{7270}$.\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 727... | [
{
"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-019599 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 10 | Use two approaches if possible: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7987} k^2\binom{7987}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain caref... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,7987\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7987(7987+1)\\cdot 2^{7985}$.\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 7987(7987+1)\\cdot 2^{7985}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-019600 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 10 | Warm-up: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3017} k\binom{3017}{k}.$$
(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... | [
{
"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 $[n]=\\{1,2,\\dots,3017\\}$. 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: $\\boxed{3017\\cdot 2^{3016}$.\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 $(1+x)^n$. |
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