id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-017401 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Solve with verification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3314} k^2\binom{3314}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully wh... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3314(3314+1)\\cdot 2^{3312}$.\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 3314(3314+1)\\cdot 2^{3312}.... | [
{
"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 $\boxed{3314(3314+1)\cdot 2^{3312}$.) |
math-017402 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Determine the requested value: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{75} k^2\binom{75}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,75\\}$ and $(a,b)\\in A\\times ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{75(75+1)\\cdot 2^{73}$.\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 75(75+1)\\cdot 2^{73}.",
"robustness_analysis": "If the... | [
{
"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-017403 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Solve with verification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5403} k^2\binom{5403}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) E... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5403(5403+1)\\cdot 2^{5401}$.\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 5403(5403+1)\\cdot 2^{5401}.... | [
{
"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 $\boxed{5403(5403+1)\cdot 2^{5401}$.) |
math-017404 | Combinatorics: Binomial Sums — Double Counting | 9 | Indicate where a theorem is used: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{65} k^2\binom{65}{k}.$$
(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": "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 $A\\subseteq[n]=\\{1,\\dots,65\\}$ and $(a,b)\\in A\\times ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{65(65+1)\\cdot 2^{63}$.\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 65(65+1)\\cdot 2^{63}.",
"robust... | [
{
"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 $\boxed{65(65+1)\cdot 2^{63}$.) |
math-017405 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Exercise: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7777} k\binom{7777}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the ... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7777\\cdot 2^{7776}$.\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 777... | [
{
"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-017406 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Work this out carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1994} k^2\binom{1994}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully wh... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1994\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1994(1994+1)\\cdot 2^{1992}$.\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 1994(1994+1)\\cdot 2^{1992}.... | [
{
"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 $\boxed{1994(1994+1)\cdot 2^{1992}$.) |
math-017407 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Proceed methodically: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6413} k\binom{6413}{k}.$$
(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 $[n]=\\{1,2,\\dots,6413\\}$. 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: $\\boxed{6413\\cdot 2^{6412}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 641... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | 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-017408 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Make each step logically reversible (or explain if not): Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3747} k^2\binom{3747}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family o... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,3747\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3747(3747+1)\\cdot 2^{3745}$.\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 3747(3747+1)\\cdot 2^{3745}.",
"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 $\boxed{3747(3747+1)\cdot 2^{3745}$.) |
math-017409 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Explain what is being counted/optimized: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{3333} k\binom{3333}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3333\\cdot 2^{3332}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{3333\cdot 2^{3332}$.) |
math-017410 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Complete the analysis: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6443} k^2\binom{6443}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6443\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6443(6443+1)\\cdot 2^{6441}$.\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 6443(6443+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 $\boxed{6443(6443+1)\cdot 2^{6441}$.) |
math-017411 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Explain what is being counted/optimized: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{245} k\binom{245}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) B... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,245\\}$. 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: $\\boxed{245\\cdot 2^{244}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | 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 $\boxed{245\cdot 2^{244}$.) |
math-017412 | Combinatorics: Binomial Sums — Double Counting | 9 | Prompt: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4389} k\binom{4389}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the sa... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,4389\\}$. 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: $\\boxed{4389\\cdot 2^{4388}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is 438... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Key idea: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{4389\cdot 2^{4388}$.) |
math-017413 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Task: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5676} k^2\binom{5676}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why your combinatoria... | [
{
"method_name": "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 $A\\subseteq[n]=\\{1,\\dots,5676\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5676(5676+1)\\cdot 2^{5674}$.\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 5676(5676+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{5676(5676+1)\cdot 2^{5674}$.) |
math-017414 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | 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
$$S=\sum_{k=0}^{5976} k\binom{5976}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{5976\\cdot 2^{5975}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-017415 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Explain what is being counted/optimized: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5646} k^2\binom{5646}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5646(5646+1)\\cdot 2^{5644}$.\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 5646(5646+1)\\cdot 2^{5644}.... | [
{
"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 $\boxed{5646(5646+1)\cdot 2^{5644}$.) |
math-017416 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | 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
$$S=\sum_{k=0}^{2515} k^2\binom{2515}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appr... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2515\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2515(2515+1)\\cdot 2^{2513}$.\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 2515(2515+1)\\cdot 2^{2513}.",
"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-017417 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Solve and then verify: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2783} k^2\binom{2783}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2783\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2783(2783+1)\\cdot 2^{2781}$.\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 2783(2783+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{2783(2783+1)\cdot 2^{2781}$.) |
math-017418 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Indicate where a theorem is used: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5156} k^2\binom{5156}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5156(5156+1)\\cdot 2^{5154}$.\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 5156(5156+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-017419 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Solve and then verify: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6491} k^2\binom{6491}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6491\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6491(6491+1)\\cdot 2^{6489}$.\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 6491(6491+1)\\cdot 2^{6489}.",
"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 $\boxed{6491(6491+1)\cdot 2^{6489}$.) |
math-017420 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | State any required conditions first: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5839} k\binom{5839}{k}.$$
(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 $[n]=\\{1,2,\\dots,5839\\}$. 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: $\\boxed{5839\\cdot 2^{5838}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-017421 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Be explicit about assumptions: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{5486} k\binom{5486}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain w... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,5486\\}$. 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: $\\boxed{5486\\cdot 2^{5485}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-017422 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | State any required conditions first: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{197} k^2\binom{197}{k}.$$
(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: $\\boxed{197(197+1)\\cdot 2^{195}$.\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 197(197+1)\\cdot 2^{195}.",
"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 $\boxed{197(197+1)\cdot 2^{195}$.) |
math-017423 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Answer with a short justification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7685} k\binom{7685}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7685\\cdot 2^{7684}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{7685\cdot 2^{7684}$.) |
math-017424 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Do not skip justification steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6636} k\binom{6636}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both a... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6636\\cdot 2^{6635}$.\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-017425 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Provide both a computational and a conceptual explanation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4254} k\binom{4254}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-coun... | [
{
"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 $[n]=\\{1,2,\\dots,4254\\}$. 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: $\\boxed{4254\\cdot 2^{4253}$.\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-017426 | Combinatorics: Binomial Sums — Double Counting | 9 | Solve (and briefly cross-validate): Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6887} k\binom{6887}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why bot... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,6887\\}$. 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: $\\boxed{6887\\cdot 2^{6886}$.\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 688... | [
{
"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 $\boxed{6887\cdot 2^{6886}$.) |
math-017427 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Do not skip justification steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1024} k^2\binom{1024}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain care... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1024\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1024(1024+1)\\cdot 2^{1022}$.\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 1024(1024+1)\\cdot 2^{1022}.",
"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-017428 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Track quantifiers carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6574} k^2\binom{6574}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,6574\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6574(6574+1)\\cdot 2^{6572}$.\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 6574(6574+1)\\cdot 2^{6572}.",
"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-017429 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Start by stating any domain restrictions: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{685} k^2\binom{685}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,685\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{685(685+1)\\cdot 2^{683}$.\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 685(685+1)\\cdot 2^{683}.",
"... | [
{
"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-017430 | Combinatorics: Binomial Sums — Double Counting | 9 | Write the solution set clearly: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4863} k^2\binom{4863}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain ca... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4863(4863+1)\\cdot 2^{4861}$.\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 4863(4863+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 $\boxed{4863(4863+1)\cdot 2^{4861}$.) |
math-017431 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Write the solution set clearly: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1610} k^2\binom{1610}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1610(1610+1)\\cdot 2^{1608}$.\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 1610(1610+1)\\cdot 2^{1608}.... | [
{
"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-017432 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Track units/moduli carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2467} k^2\binom{2467}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain careful... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2467\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2467(2467+1)\\cdot 2^{2465}$.\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 2467(2467+1)\\cdot 2^{2465}.",
"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 $\boxed{2467(2467+1)\cdot 2^{2465}$.) |
math-017433 | Combinatorics: Binomial Sums — Double Counting | 9 | Find the exact value: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4852} k^2\binom{4852}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain care... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4852\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4852(4852+1)\\cdot 2^{4850}$.\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 4852(4852+1)\\cdot 2^{4850}.",
"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 $\boxed{4852(4852+1)\cdot 2^{4850}$.) |
math-017434 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Complete the analysis: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{805} k\binom{805}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ap... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,805\\}$. 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: $\\boxed{805\\cdot 2^{804}$.\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 805\\... | [
{
"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 $\boxed{805\cdot 2^{804}$.) |
math-017435 | Combinatorics: Binomial Sums — Double Counting | 9 | Track units/moduli carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5245} k^2\binom{5245}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain care... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,5245\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5245(5245+1)\\cdot 2^{5243}$.\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 5245(5245+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 $\boxed{5245(5245+1)\cdot 2^{5243}$.) |
math-017436 | Combinatorics: Binomial Sums — Double Counting | 9 | Question: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2529} k\binom{2529}{k}.$$
(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 $[n]=\\{1,2,\\dots,2529\\}$. 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: $\\boxed{2529\\cdot 2^{2528}$.\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-017437 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Give reasoning, not just computation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{6818} k^2\binom{6818}{k}.$$
(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: $\\boxed{6818(6818+1)\\cdot 2^{6816}$.\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 6818(6818+1)\\cdot 2^{6816}.",
"robustness_analys... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Takeaway: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{6818(6818+1)\cdot 2^{6816}$.) |
math-017438 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Compute the requested quantity: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2036} k\binom{2036}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ap... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,2036\\}$. 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: $\\boxed{2036\\cdot 2^{2035}$.\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 $\boxed{2036\cdot 2^{2035}$.) |
math-017439 | Combinatorics: Binomial Sums — Double Counting | 9 | Exercise: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6567} k^2\binom{6567}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why yo... | [
{
"method_name": "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 $A\\subseteq[n]=\\{1,\\dots,6567\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6567(6567+1)\\cdot 2^{6565}$.\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 6567(6567+1)\\cdot 2^{6565}.",
"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-017440 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Give an answer and a quick verification: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5651} k^2\binom{5651}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family o... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,5651\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5651(5651+1)\\cdot 2^{5649}$.\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 5651(5651+1)\\cdot 2^{5649}.",
"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-017441 | Combinatorics: Binomial Sums — Double Counting | 9 | Explain what is being counted/optimized: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2434} k\binom{2434}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain wh... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,2434\\}$. 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: $\\boxed{2434\\cdot 2^{2433}$.\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 243... | [
{
"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-017442 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Give reasoning, not just computation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{940} k\binom{940}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Brie... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{940\\cdot 2^{939}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | 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-017443 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Explain each transformation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6587} k\binom{6587}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6587\\cdot 2^{6586}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-017444 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Carefully track domains: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5505} k^2\binom{5505}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5505(5505+1)\\cdot 2^{5503}$.\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 5505(5505+1)\\cdot 2^{5503}.",
"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 $\boxed{5505(5505+1)\cdot 2^{5503}$.) |
math-017445 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Solve (and briefly cross-validate): Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2597} k^2\binom{2597}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of tri... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,2597\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2597(2597+1)\\cdot 2^{2595}$.\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 2597(2597+1)\\cdot 2^{2595}.",
"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 $\boxed{2597(2597+1)\cdot 2^{2595}$.) |
math-017446 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Give reasoning, not just computation: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7247} k\binom{7247}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why b... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,7247\\}$. 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: $\\boxed{7247\\cdot 2^{7246}$.\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-017447 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Checkpoint: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{1710} k^2\binom{1710}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why your combin... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1710\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1710(1710+1)\\cdot 2^{1708}$.\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 1710(1710+1)\\cdot 2^{1708}.",
"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 $\boxed{1710(1710+1)\cdot 2^{1708}$.) |
math-017448 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Give an answer and a quick verification: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3878} k^2\binom{3878}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Expl... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3878(3878+1)\\cdot 2^{3876}$.\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 3878(3878+1)\\cdot 2^{3876}.",
"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 $\boxed{3878(3878+1)\cdot 2^{3876}$.) |
math-017449 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | State any required conditions first: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4503} k\binom{4503}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Bri... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4503\\cdot 2^{4502}$.\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 $\boxed{4503\cdot 2^{4502}$.) |
math-017450 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Solve and justify each step: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4798} k\binom{4798}{k}.$$
(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: $\\boxed{4798\\cdot 2^{4797}$.\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 479... | [
{
"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 $\boxed{4798\cdot 2^{4797}$.) |
math-017451 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Keep the final answer in boxed form: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2642} k\binom{2642}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why bo... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,2642\\}$. 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: $\\boxed{2642\\cdot 2^{2641}$.\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 264... | [
{
"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 $\boxed{2642\cdot 2^{2641}$.) |
math-017452 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Explain why your operations are valid: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4996} k\binom{4996}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly e... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4996\\cdot 2^{4995}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-017453 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Provide a rigorous solution: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{6601} k^2\binom{6601}{k}.$$
(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: $\\boxed{6601(6601+1)\\cdot 2^{6599}$.\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 6601(6601+1)\\cdot 2^{6599}.",
"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-017454 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Keep the final answer in boxed form: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{638} k^2\binom{638}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{638(638+1)\\cdot 2^{636}$.\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 638(638+1)\\cdot 2^{63... | [
{
"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 $\boxed{638(638+1)\cdot 2^{636}$.) |
math-017455 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Give a fully justified solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1789} k\binom{1789}{k}.$$
(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 $[n]=\\{1,2,\\dots,1789\\}$. 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: $\\boxed{1789\\cdot 2^{1788}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Core principle: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{1789\cdot 2^{1788}$.) |
math-017456 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Make each step logically reversible (or explain if not): Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6762} k\binom{6762}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argum... | [
{
"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 $[n]=\\{1,2,\\dots,6762\\}$. 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: $\\boxed{6762\\cdot 2^{6761}$.\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 $\boxed{6762\cdot 2^{6761}$.) |
math-017457 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Indicate where a theorem is used: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1612} k\binom{1612}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why bo... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1612\\cdot 2^{1611}$.\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 $\boxed{1612\cdot 2^{1611}$.) |
math-017458 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Give a fully justified solution: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{692} k^2\binom{692}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,692\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{692(692+1)\\cdot 2^{690}$.\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 692(692+1)\\cdot 2^{69... | [
{
"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-017459 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Track quantifiers carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{7830} k\binom{7830}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both appro... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,7830\\}$. 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: $\\boxed{7830\\cdot 2^{7829}$.\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 783... | [
{
"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 $\boxed{7830\cdot 2^{7829}$.) |
math-017460 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Complete the analysis: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5059} k\binom{5059}{k}.$$
(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 $[n]=\\{1,2,\\dots,5059\\}$. 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: $\\boxed{5059\\cdot 2^{5058}$.\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 $\boxed{5059\cdot 2^{5058}$.) |
math-017461 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Problem: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{222} k^2\binom{222}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why your combinat... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,222\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{222(222+1)\\cdot 2^{220}$.\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 222(222+1)\\cdot 2^{220}.",
"... | [
{
"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-017462 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Work carefully and justify each inference: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1311} k\binom{1311}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1311\\cdot 2^{1310}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{1311\cdot 2^{1310}$.) |
math-017463 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Indicate where a theorem is used: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{7344} k\binom{7344}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefl... | [
{
"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: $\\boxed{7344\\cdot 2^{7343}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-017464 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Solve and justify each step: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{6192} k^2\binom{6192}{k}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{6192(6192+1)\\cdot 2^{6190}$.\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 6192(6192+1)\\cdot 2^{6190}.",
"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 $\boxed{6192(6192+1)\cdot 2^{6190}$.) |
math-017465 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Find the exact value: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{4673} k^2\binom{4673}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully wh... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,4673\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4673(4673+1)\\cdot 2^{4671}$.\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 4673(4673+1)\\cdot 2^{4671}.... | [
{
"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-017466 | Combinatorics: Binomial Sums — Double Counting | 9 | Solve with verification: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7287} k\binom{7287}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approa... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,7287\\}$. 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: $\\boxed{7287\\cdot 2^{7286}$.\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 $\boxed{7287\cdot 2^{7286}$.) |
math-017467 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Write the solution set clearly: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{7492} k^2\binom{7492}{k}.$$
(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: $\\boxed{7492(7492+1)\\cdot 2^{7490}$.\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 7492(7492+1)\\cdot 2^{7490}.",
"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 $\boxed{7492(7492+1)\cdot 2^{7490}$.) |
math-017468 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Do not skip justification steps: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{7433} k\binom{7433}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why bot... | [
{
"method_name": "Double Counting (Subset + Distinguished Element)",
"approach": "Count pairs $(A,a)$ where $A\\subseteq[n]$ and $a\\in A$ two different ways.",
"steps": [
"Step 1: Let $[n]=\\{1,2,\\dots,7433\\}$. 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: $\\boxed{7433\\cdot 2^{7432}$.\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 743... | [
{
"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-017469 | Combinatorics: Binomial Sums — Double Counting | 9 | Warm-up: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{945} k\binom{945}{k}.$$
(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 q... | [
{
"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: $\\boxed{945\\cdot 2^{944}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here is... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. |
math-017470 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Carefully track domains: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{1497} k^2\binom{1497}{k}.$$
(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: $\\boxed{1497(1497+1)\\cdot 2^{1495}$.\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 1497(1497+1)\\cdot 2^{1495}.... | [
{
"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 $\boxed{1497(1497+1)\cdot 2^{1495}$.) |
math-017471 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Prompt: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3306} k^2\binom{3306}{k}.$$
(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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3306(3306+1)\\cdot 2^{3304}$.\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 3306(3306+1)\\cdot 2^{3304}.... | [
{
"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 $\boxed{3306(3306+1)\cdot 2^{3304}$.) |
math-017472 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Provide both a computational and a conceptual explanation: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{4048} k\binom{4048}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting arg... | [
{
"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: $\\boxed{4048\\cdot 2^{4047}$.\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 404... | [
{
"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 $\boxed{4048\cdot 2^{4047}$.) |
math-017473 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | 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
$$S=\sum_{k=0}^{3505} k^2\binom{3505}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appr... | [
{
"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: $\\boxed{3505(3505+1)\\cdot 2^{3503}$.\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 3505(3505+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 $\boxed{3505(3505+1)\cdot 2^{3503}$.) |
math-017474 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Challenge: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{380} k\binom{380}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches count the s... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{380\\cdot 2^{379}$.\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 380\\... | [
{
"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-017475 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Warm-up: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{823} k^2\binom{823}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully w... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,823\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{823(823+1)\\cdot 2^{821}$.\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 823(823+1)\\cdot 2^{821}.",
"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-017476 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Task: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{189} k^2\binom{189}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,189\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{189(189+1)\\cdot 2^{187}$.\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 189(189+1)\\cdot 2^{187}.",
"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 $\boxed{189(189+1)\cdot 2^{187}$.) |
math-017477 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Do not skip justification steps: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5177} k^2\binom{5177}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain care... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,5177\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5177(5177+1)\\cdot 2^{5175}$.\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 5177(5177+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-017478 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Solve and sanity-check: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3061} k\binom{3061}{k}.$$
(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 $[n]=\\{1,2,\\dots,3061\\}$. 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: $\\boxed{3061\\cdot 2^{3060}$.\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 $\boxed{3061\cdot 2^{3060}$.) |
math-017479 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Challenge: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{3123} k^2\binom{3123}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why y... | [
{
"method_name": "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: $\\boxed{3123(3123+1)\\cdot 2^{3121}$.\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 3123(3123+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 $\boxed{3123(3123+1)\cdot 2^{3121}$.) |
math-017480 | Combinatorics: Binomial Sums — Double Counting | 9 | Solve and then verify: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{243} k\binom{243}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both ap... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{243\\cdot 2^{242}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | 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-017481 | Combinatorics: Binomial Sums — Double Counting | 9 | Solve and then verify: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3581} k\binom{3581}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches ... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3581\\cdot 2^{3580}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1}$, which here ... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Remember: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{3581\cdot 2^{3580}$.) |
math-017482 | Combinatorics: Binomial Sums — Double Counting | 9 | Provide a rigorous solution: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2168} k\binom{2168}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both appro... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2168\\cdot 2^{2167}$.\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-017483 | Combinatorics: Binomial Sums — Double Counting | 9 | Where appropriate, name the theorem you use: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4510} k\binom{4510}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explai... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4510\\cdot 2^{4509}$.\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 $\boxed{4510\cdot 2^{4509}$.) |
math-017484 | Combinatorics: Binomial Sums — Double Counting | 9 | Track units/moduli carefully: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{4918} k^2\binom{4918}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain careful... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4918(4918+1)\\cdot 2^{4916}$.\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 4918(4918+1)\\cdot 2^{4916}.... | [
{
"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-017485 | Combinatorics: Binomial Sums — Double Counting | 9 | Question: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{663} k\binom{663}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approaches coun... | [
{
"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: $\\boxed{663\\cdot 2^{662}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yield $n2^{n-1... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | 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-017486 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Carefully track domains: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4707} k^2\binom{4707}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) E... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4707(4707+1)\\cdot 2^{4705}$.\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 4707(4707+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 $\boxed{4707(4707+1)\cdot 2^{4705}$.) |
math-017487 | Combinatorics: Binomial Sums — Double Counting | 9 | Answer with a short justification: Find a closed form for the sum and explicitly state what combinatorial objects it counts:
Compute the sum
$$S=\sum_{k=0}^{776} k\binom{776}{k}.$$
(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 $[n]=\\{1,2,\\dots,776\\}$. 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: $\\boxed{776\\cdot 2^{775}$.\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 776\\... | [
{
"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 $\boxed{776\cdot 2^{775}$.) |
math-017488 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Compute the requested quantity: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{4546} k^2\binom{4546}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4546(4546+1)\\cdot 2^{4544}$.\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 4546(4546+1)\\cdot 2^{4544}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{4546(4546+1)\cdot 2^{4544}$.) |
math-017489 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Find the exact value: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{5628} k^2\binom{5628}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why y... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,5628\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5628(5628+1)\\cdot 2^{5626}$.\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 5628(5628+1)\\cdot 2^{5626}.... | [
{
"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-017490 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Indicate where a theorem is used: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{2445} k^2\binom{2445}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,2445\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2445(2445+1)\\cdot 2^{2443}$.\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 2445(2445+1)\\cdot ... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
math-017491 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Explain why your operations are valid: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{5651} k\binom{5651}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain w... | [
{
"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: $\\boxed{5651\\cdot 2^{5650}$.\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 565... | [
{
"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 $\boxed{5651\cdot 2^{5650}$.) |
math-017492 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Solve and sanity-check: Compute the sum and reconcile the algebraic and combinatorial interpretations:
Compute the sum
$$S=\sum_{k=0}^{3512} k^2\binom{3512}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully why... | [
{
"method_name": "Operator Method: $(x\\frac{d}{dx})^2$",
"approach": "The operator $T=x\\frac{d}{dx}$ multiplies the coefficient of $x^k$ by $k$; applying it twice produces $k^2$.",
"steps": [
"Step 1: Start with $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Apply $T=x\\frac{d}{dx}$ ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3512(3512+1)\\cdot 2^{3510}$.\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 3512(3512+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-017493 | Discrete Math: Operators — $x\frac{d}{dx}$ Trick | 9 | Derive the result step-by-step: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{6406} k^2\binom{6406}{k}.$$
(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: $\\boxed{6406(6406+1)\\cdot 2^{6404}$.\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 6406(6406+1)\\cdot 2^{6404}.",
"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 $\boxed{6406(6406+1)\cdot 2^{6404}$.) |
math-017494 | Combinatorics: Binomial Sums — Double Counting | 9 | Provide a rigorous solution: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{2979} k\binom{2979}{k}.$$
(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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2979\\cdot 2^{2978}$.\nBoth methods compute the same count of pairs $(A,a)$: differentiation produces the weighted sum, while double counting counts the same pairs by choosing $a$ first. Both yie... | [
{
"error_description": "Differentiated but forgot to multiply by $x$ before setting $x=1$.",
"why_plausible": "The exponent shift $x^{k-1}$ is easy to miss.",
"why_wrong": "Without multiplying by $x$, the series is $\\sum k\\binom{n}{k}x^{k-1}$, which is not the target sum.",
"which_method_catches_i... | Takeaway: Weights like $k\binom{n}{k}$ usually mean 'choose a $k$-subset and then choose a distinguished element inside it'; algebraically the same weight arises from differentiating $(1+x)^n$. (Here the result is $\boxed{2979\cdot 2^{2978}$.) |
math-017495 | Combinatorics: Binomial Moments — $\sum k\binom{n}{k}$ | 9 | Work this out carefully: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{666} k\binom{666}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both approach... | [
{
"method_name": "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 $[n]=\\{1,2,\\dots,666\\}$. 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: $\\boxed{666\\cdot 2^{665}$.\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... | 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-017496 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Task: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{1010} k^2\binom{1010}{k}.$$
(a) Solve using generating functions (apply $x\frac{d}{dx}$ twice).
(b) Solve by double counting an appropriate family of triples.
(c) Explain carefully wh... | [
{
"method_name": "Double Counting (Subset + Ordered Pair in Subset)",
"approach": "Count triples $(A,a,b)$ where $A\\subseteq[n]$ and $(a,b)\\in A\\times A$ (ordered, repetition allowed).",
"steps": [
"Step 1: Count triples $(A,a,b)$ with $A\\subseteq[n]=\\{1,\\dots,1010\\}$ and $(a,b)\\in A\\time... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1010(1010+1)\\cdot 2^{1008}$.\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 1010(1010+1)\\cdot 2^{1008}.",
"robustness_... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. (Here the result is $\boxed{1010(1010+1)\cdot 2^{1008}$.) |
math-017497 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Start by stating any domain restrictions: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{605} k^2\binom{605}{k}.$$
(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 $A\\subseteq[n]=\\{1,\\dots,605\\}$ and $(a,b)\\in A\\times... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{605(605+1)\\cdot 2^{603}$.\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 605(605+1)\\cdot 2^{60... | [
{
"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 $\boxed{605(605+1)\cdot 2^{603}$.) |
math-017498 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Work carefully and justify each inference: Evaluate the sum. One method must use $(1+x)^n$; the other must be combinatorial:
Compute the sum
$$S=\sum_{k=0}^{1771} k\binom{1771}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly expla... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1771\\cdot 2^{1770}$.\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-017499 | Combinatorics: Binomial Sums — Differentiating Generating Functions | 9 | Question: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{3247} k\binom{3247}{k}.$$
(a) Solve using a generating-function/differentiation argument.
(b) Solve by a combinatorial double-counting argument.
(c) Briefly explain why both appro... | [
{
"method_name": "Differentiate the Binomial Theorem",
"approach": "Differentiate $(1+x)^n$ and evaluate at $x=1$ to create the weight $k$ on $\\binom{n}{k}$.",
"steps": [
"Step 1: $(1+x)^n=\\sum_{k=0}^n \\binom{n}{k}x^k$.",
"Step 2: Differentiate: $n(1+x)^{n-1}=\\sum_{k=0}^n k\\binom{n}{k}x... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3247\\cdot 2^{3246}$.\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 $\boxed{3247\cdot 2^{3246}$.) |
math-017500 | Combinatorics: Binomial Moments — $\sum k^2\binom{n}{k}$ | 9 | Explain each transformation: Compute the binomial sum using (i) generating functions/differentiation and (ii) double counting:
Compute the sum
$$S=\sum_{k=0}^{5377} k^2\binom{5377}{k}.$$
(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: $\\boxed{5377(5377+1)\\cdot 2^{5375}$.\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 5377(5377+1)\\cdot 2^{5375}.... | [
{
"error_description": "Modeled $k^2$ as choosing two distinct elements and used $\\binom{k}{2}$.",
"why_plausible": "Both involve 'two elements from a $k$-set', so confusion is common.",
"why_wrong": "$k^2$ counts ordered pairs with repetition; $\\binom{k}{2}$ counts unordered distinct pairs.",
"wh... | Core principle: Higher-moment binomial sums correspond to counting subsets with extra structure (like ordered pairs inside the subset). Algebraically, $x\frac{d}{dx}$ is the operator that inserts the weight $k$. |
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