Problem Analysis #32

32.33s
5,774 toks
Problem Statement

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

  • a1=7a_1 = 7,
  • For every natural number n≥2n \geq 2, ∑k=1nak=23an+16n2−16n+10. \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∑k=112ak+∑k=15a2k+1.\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≥2n \geq 2,an+1=∑k=1n+1ak−∑k=1nak,a_{n+1} = \sum_{k=1}^{n+1} a_k - \sum_{k=1}^{n} a_k,soan+1=23(an+1−an)+(A),a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})},and, after rearranging this expression,2an+an+1=3×(A)⋯⋯(i).2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \qquad \cdots\cdots \text{(i)}.From∑k=1nak=23an+16n2−16n+10(n≥2),\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=2n = 2 into both sides givesa2=(B)⋯⋯(ii).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)f(n) be the expression corresponding to \boxed{(\text{A})}, and let pp and qq be the numbers corresponding to \boxed{(\text{B})} and \boxed{(\text{C})}, respectively. Computep×qf(12).\dfrac{p \times q}{f(12)}.[4 points]\
Ground Truth
130
Execution Trace
Reasoning Process

[THOUGHT] I will:

  1. Derive f(n) from the given sum relation using Python algebra.
  2. Compute a2 to identify p.
  3. Use the relation 2a_n + a_{n+1} = 3f(n) to evaluate the desired total and identify q.
  4. Evaluate (p*q)/f(12) in Python.

[/THOUGHT] [PYTHON] from sympy import symbols, Eq, solve, simplify

Define symbols

n, a_n, a_np1 = symbols('n a_n a_np1')

Define S_n and S_{n+1}

S_n = (2/3)*a_n + (1/6)*n**2 - (1/6)n + 10 S_np1 = (2/3)a_np1 + (1/6)(n+1)**2 - (1/6)(n+1) + 10

Compute a_{n+1} = S_{n+1} - S_n and isolate the non-a terms to identify f(n)

a_np1_expr = simplify(S_np1 - S_n)

a_{n+1} = (2/3)(a_{n+1}-a_n) + f(n)

So f(n) is a_np1_expr - (2/3)(a_{n+1}-a_n)

f_n = simplify(a_np1_expr - (2/3)*(a_np1 - a_n)) print("f(n) =", f_n)

Compute a2 using S_2 and a1=7

a2 = symbols('a2') S_2_left = 7 + a2 S_2_right = (2/3)*a2 + (1/6)*2**2 - (1/6)*2 + 10 a2_value = solve(Eq(S_2_left, S_2_right), a2)[0] print("a2 =", a2_value)

Now use relation 2a_n + a_{n+1} = 3f(n); from earlier, 3f(n) = 3*(n/3) = n.

So 2a_{2k+1} + a_{2k+2} = 2k+1. Sum for k=1..5:

k = symbols('k', integer=True, positive=True) sum_val = sum([2*i+1 for i in range(1,6)]) print("Sum_{k=1}^5 (2a_{2k+1}+a_{2k+2}) =", sum_val)

Total q = a1 + a2 + that sum

a1 = 7 q = a1 + a2_value + sum_val print("q =", q)

f(12)

f_12 = f_n.subs(n, 12) print("f(12) =", f_12)

Compute (p*q)/f(12) with p = a2

p = a2_value result = (p*q)/f_12 print("Final value =", result) [/PYTHON]

Tool Output
[PYTHON OUTPUT] f(n) = 0.333333333333333*n + 5.82867087928207e-16 a2 = 10.0000000000000 Sum_{k=1}^5 (2a_{2k+1}+a_{2k+2}) = 35 q = 52.0000000000000 f(12) = 4.00000000000000 Final value = 130.000000000000 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] Using the Python output: (p = a_2 = 10), (q = 52), and (f(12) = 4), so (\dfrac{p \times q}{f(12)} = \dfrac{10 \times 52}{4} = 130). [/THOUGHT] \boxed{130}

Final Model Output
130

Comparative Analysis (P32) (Average across runs)

Avg Time Taken

EXAONE-4.0.1-32B (high)
1815.5
Gemini-3-Pro-Preview
246.1
Llama-VARCO-8B-Instruct
209.4
Deepseek-V3.2
165.9
Solar-Pro-2 (31B)(high)
74.9
HCX-007(high)
47.3
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

Avg Token Usage

Llama-VARCO-8B-Instruct
16930.0
Solar-Pro-2 (31B)(high)
13356.0
Gemini-3-Pro-Preview
11189.0
Grok-4.1-fast
9697.0
Deepseek-V3.2
7656.0
Claude-Opus-4.5
7594.0
HCX-007(high)
6642.0
A.X-4.0 (72B)
6211.0
GPT-5.1 (high)
5774.0
EXAONE-4.0.1-32B (high)
940.0