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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 20 Let{a n}be a sequence satisfying the following conditions. •a 1 = 7, •For every natural numbern≥2, nX k=1 ak = 2 3 an + 1 6 n2 − 1 6 n+ 10. Consider the following work to compute 12X k=1 ak + 5X k=1 a2k+1. 7 For every natural numbern≥2, an+1 = n+1X k=1 ak − nX k=1 ak, so an+1 = 2 3(an+1 −a n) + (A) , and, after rearranging this expression, 2an +a n+1 = 3× (A) · · · · · ·(i). From nX k=1 ak = 2 3 an + 1 6 n2 − 1 6 n+ 10 (n≥2), substitutingn= 2 into both sides gives a2 = (B) · · · · · ·(ii). By (i) and (ii), 12X k=1 ak + 5X k=1 a2k+1 =a 1 +a 2 + 5X k=1 (2a2k+1 +a 2k+2) = (C) . Letf(n) be the expression corresponding to (A) , and letpandqbe the numbers corresponding to (B) and (C) , respectively. Compute p×q f(12) . [4 points] 더보기

Looking at this problem, I need to find (A), (B), (C), then compute p×q/f(12).

Finding (A): For n≥2, using the sum formula at n and n+1:

𝑎 𝑛 + 1

∑ 𝑘

1 𝑛 + 1 𝑎 𝑘 − ∑ 𝑘

1 𝑛 𝑎 𝑘

2 3 𝑎 𝑛 + 1 − 2 3 𝑎 𝑛 + 𝑛 3 a n+1 ​

=∑ k=1 n+1 ​

a k ​

−∑ k=1 n ​

a k ​

= 3 2 ​

a n+1 ​

− 3 2 ​

a n ​

3 n ​

So 𝑎 𝑛 + 1

2 3 ( 𝑎 𝑛 + 1 − 𝑎 𝑛 ) + 𝑛 3 a n+1 ​

= 3 2 ​

(a n+1 ​

−a n ​

)+ 3 n ​

, giving (A) = n/3, thus 𝑓 ( 𝑛 )

𝑛 3 f(n)= 3 n ​

.

Rearranging: 2 𝑎 𝑛 + 𝑎 𝑛 + 1

𝑛 2a n ​

+a n+1 ​

=n for n≥2.

Finding (B): Substituting n=2 into the sum formula:

𝑎 1 + 𝑎 2

2 3 𝑎 2 + 1 6 ( 4 ) − 1 6 ( 2 ) + 10 a 1 ​

+a 2 ​

= 3 2 ​

a 2 ​

6 1 ​

(4)− 6 1 ​

(2)+10

7 + 𝑎 2

2 3 𝑎 2 + 32 3 7+a 2 ​

= 3 2 ​

a 2 ​

3 32 ​

1 3 𝑎 2

11 3 ⇒ 𝑎 2

11 3 1 ​

a 2 ​

= 3 11 ​

⇒a 2 ​

=11

So p = (B) = 11.

Finding (C):

∑ 𝑘

1 12 𝑎 𝑘 + ∑ 𝑘

1 5 𝑎 2 𝑘 + 1

𝑎 1 + 𝑎 2 + ∑ 𝑘

1 5 ( 2 𝑎 2 𝑘 + 1 + 𝑎 2 𝑘 + 2 ) ∑ k=1 12 ​

a k ​

+∑ k=1 5 ​

a 2k+1 ​

=a 1 ​

+a 2 ​

+∑ k=1 5 ​

(2a 2k+1 ​

+a 2k+2 ​

)

Using 2 𝑎 𝑛 + 𝑎 𝑛 + 1

𝑛 2a n ​

+a n+1 ​

=n:

k=1: 2a₃+a₄ = 3 k=2: 2a₅+a₆ = 5 k=3: 2a₇+a₈ = 7 k=4: 2a₉+a₁₀ = 9 k=5: 2a₁₁+a₁₂ = 11

Sum = 3+5+7+9+11 = 35

So (C) = 7 + 11 + 35 = 53, thus q = 53.

Computing: f(12) = 12/3 = 4

𝑝 × 𝑞 𝑓 ( 12 )

11 × 53 4

583 4 f(12) p×q ​

= 4 11×53 ​

= 4 583 ​

\boxed{\dfrac{583}{4}}

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