Problem Analysis #32

EXAONE-4.0.1-32B (high) · Run 1

1815.48s

940 toks

Problem Statement

Let $\{a_n\}$ be a sequence satisfying the following conditions.\begin{center}\fbox{\begin{minipage}{0.9\textwidth}

$a_1 = 7$ ,
For every natural number $n \geq 2$ ,

\sum_{k=1}^{n} a_k = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10.

\end{minipage}}\end{center}Consider the following work to compute

\sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1}.

\begin{center}\fbox{\begin{minipage}{0.9\textwidth}For every natural number $n \geq 2$ ,

a_{n+1} = \sum_{k=1}^{n+1} a_k - \sum_{k=1}^{n} a_k,

a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})},

and, after rearranging this expression,

2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \qquad \cdots\cdots \text{(i)}.

From

\sum_{k=1}^{n} a_k = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10 \quad (n \geq 2),

substituting $n = 2$ into both sides gives

a_2 = \boxed{(\text{B})} \qquad \cdots\cdots \text{(ii)}.

By (i) and (ii),

\sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} = a_1 + a_2 + \sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2})

= \boxed{(\text{C})}.

\end{minipage}}\end{center}Let $f(n)$ be the expression corresponding to $\boxed{(\text{A})}$ , and let $p$ and $q$ be the numbers corresponding to $\boxed{(\text{B})}$ and $\boxed{(\text{C})}$ , respectively. Compute

\dfrac{p \times q}{f(12)}.

[4 points]\

Ground Truth

130

Execution Trace

Reasoning Process

[THOUGHT] To solve this problem, we need to follow these steps: 1. Find the expression for $f(n)$ corresponding to $\boxed{(\text{A})}$ . This comes from the recurrence relation derived in step (i):

a_{n+1} = \frac{2}{3}(a_n - a_{n-1}) + (\dots),

but we need more information. From equation $\sum k=^{m}}$ and $n$ terms, the recurrence relation can be derived as follows: Given that for every natural number $m $$\geq$$2$ ,

S_k = a_{n} -a(1/3)\cdot(\text{terms})

However from step (i), we see it's actually $5$ points. We need to compute the recurrence relation properly: From equation $\sum k=^{m}}$ . The difference between consecutive sums gives us a way forward, but let me re-express this carefully using Python code for clarity and accuracy [PYTHON]

Final Model Output

(\text{A})

Comparative Analysis (P32) (Average across runs)

Avg Time Taken

EXAONE-4.0.1-32B (high)

1815.5

K-EXAONE-236B-A23B

1469.7

Gemini-3-Pro-Preview

246.1

Llama-VARCO-8B-Instruct

209.4

Deepseek-V3.2

165.9

Solar-Open-100B

78.9

Solar-Pro-2 (31B)(high)

74.9

Kanana-2-30B-Thinking-2601

73.7

Kanana-2-30B-Thinking

65.0

HCX-007(high)

47.3

K-EXAONE-236B-A23B

33.2

A.X-4.0 (72B)

33.1

Grok-4.1-fast

32.4

GPT-5.1 (high)

32.3

Claude-Opus-4.5

30.3

GPT-5.2 (high)

23.6

Avg Token Usage

K-EXAONE-236B-A23B

172,401

Llama-VARCO-8B-Instruct

16,930

Solar-Open-100B

15,251

Kanana-2-30B-Thinking-2601

14,849

Solar-Pro-2 (31B)(high)

13,356

Kanana-2-30B-Thinking

13,274

Gemini-3-Pro-Preview

11,189

Grok-4.1-fast

9,697

Deepseek-V3.2

7,656

Claude-Opus-4.5

7,594

HCX-007(high)

6,642

K-EXAONE-236B-A23B

6,562

A.X-4.0 (72B)

6,211

GPT-5.1 (high)

5,774

GPT-5.2 (high)

5,721

EXAONE-4.0.1-32B (high)

940