Problem Analysis #32

501.86s
63,604 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

To solve the problem, we begin by analyzing the given sequence condition:

∑k=1nak=23an+16n2−16n+10for n≥2.\sum_{k=1}^{n} a_k = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10 \quad \text{for } n \geq 2.

Let us denote Sn=∑k=1nakS_n = \sum_{k=1}^n a_k. Then the recurrence becomes:

Sn=23an+16n2−16n+10for n≥2.S_n = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10 \quad \text{for } n \geq 2.

We are also given that for every natural number n≥2n \geq 2, the following holds:

an+1=Sn+1−Sn,a_{n+1} = S_{n+1} - S_n,

and from this, it is claimed that:

an+1=23(an+1−an)+(A).a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})}.

Step 1: Derive the expression in (A)

We compute an+1=Sn+−Sna_{n+1} = S_{n+} - S_n:

an+1=(23an+1+16(n+1)2−16(n+1)+10)−(23an+16n2−(B)+10).a_{n+1} = \left( \frac{2}{3}a_{n+1} + \frac{1}{6}(n+1)^2 - \frac{1}{6}(n+1) + 10 \right) - \left( \frac{2}{3}a_n + \frac{1}{6}n^2 - \boxed{(B)} + 10 \right).

Now simplify the difference:

an+1=23(an+1−an)+[16(n+1)2−16(n+1)−16n2+16n].a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \left[ \frac{1}{6}(n+1)^2 - \frac{1}{6}(n+1) - \frac{1}{6}n^2 + \frac{1}{6}n \right].

Compute the polynomial part:

16[(n+1)2−(n+1)−n2+n]=16[n2+2n+1−n−1−n2+n]=16[2n]=13n.\frac{1}{6}[(n+1)^2 - (n+1) - n^2 + n] = \frac{1}{6}[n^2 + 2n + 1 - n - 1 - n^2 + n] = \frac{1}{6}[2n] = \frac{1}{3}n.

Wait — let's double-check:

(n+1)2=n2+2n+1,so (n+1)2−(n+1)=n2+2n+1−n−1=n2+n.(n+1)^2 = n^2 + 2n + 1, \quad \text{so } (n+1)^2 - (n+1) = n^2 + 2n + 1 - n - 1 = n^2 + n.

Then:

(n^2 + n) - n^2 = n. $$ 핑계: Wait, actually, the original polynomial term is:

\frac{1}{6}(n+1)^2 - \frac{1}{6}(n+1) - \frac{1}{6}n^2 + \frac{1}{6}n.

Socompute: So compute:

\frac{1}{6}[(n+1)^2 - (n+1) - n^2 + n] = \frac{1}{6}[n^2 + 2n + 1 - n - 1 - n^2 + n] = \frac{1}{6}[2n] = \frac{1}{3}n.

Yes!Sotheentireexpressionsimplifiesto: Yes! So the entire expression simplifies to:

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

Wait, but the problem says: > $ a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})} $ Wait, no — looking back at the work: > $ a_{n+1} = \sum_{k=1}^{n+1} a_k - \sum_{k=1}^n a_k = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})} $ Ah, so they're saying that the right-hand side of the equation for $ a_{n+1} $ equals $ \frac{2}{3}(a_{n+1} - a_n) + \text{somethingthing} $. Wait, no: $ a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \text{somethingthing} $. Wait, let's re-express the difference: $ S_{n+1} - S_n = \frac{2}{3}a_{n+1} + P(n+1) - \left( \frac{2}{3}a_n + P(n) \right) $, where $ P(n) = \frac{1}{6}n^2 - \frac{1}{6}n + 10 $. Thus, $ a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + [P(n+1) - P(n)] $. Therefore, \( \boxed{(\text{A})} = P(n+1) - P(n) \). Compute \( P(n+1) - P(n) \): \( \frac{1}{6}(n+1)^2 - \frac{1}{6}(n+1) + 10 - \left( \frac{1}{6}n^2 - \frac{1}{6}n + 10 \right) \) = \( \frac{1}{6}(n^2 + 2n + 1) - \frac{1}{6}(n + 1) - \frac{1}{6}n^2 + \frac{1}{6}n \) = \( \frac{1}{6}[n^2 + 2n + 1 - n - 1 - n^2 + n] \) = \( \frac{1}{6}[2n] = \frac{1}{3}n \). Yes! Because: \( (n+1)^2 - n^2 = 2n + 1 \) \( - (n+1) + n = -1 \) So total: \( \frac{1}{6}[(2n + 1) - 1] = \frac{1}{6}(2n) = \frac{1}{3}n \). Yes, that's correct. So \( P(n+1) - P(n) = \frac{1}{3}n \). Therefore, box (A) is \( \frac{1}{3}n \). Now, moving on. The next part says: > and, after rearranging this expression,

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

Wait, let's see. From the previous step: \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \frac{1}{3}n \) Wait, no, in the work, it's written as: > \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})} \) Wait, let's check the exact wording: > "so $a_{n+1} = \sum_{k=1}^{n+1} a_k - \sum_{k=1}^{n} a_k$, so $a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})}$" Wait, that seems odd. Maybe it's a typo in the problem statement. Wait, no, the sum is \( \sum_{k=1}^{n+1} a_k = \frac{2}{3}a_{n+1} + P(n+1) \), and \( \sum_{k=1}^n a_k = \frac{2}{3}a_n + P(n) \), so subtracting: \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + [P(n+1) - P(n)] \). Yes, because \( \sum_{k=1}^{n+1} a_k - \sum_{k=1}^n a_k = a_{n+1} \), and \( \frac{2}{3}a_{n+1} + P(n+1) - \left( \frac{2}{3}a_n + P(n) \right) = \frac{2}{3}(a_{n+1} - a_n) + (P(n+1) - P(n)) \). Therefore, equating the two expressions for \( a_{n+1} \): \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + (P(n+1) - P(n)) \) Then, moving terms around: \( a_{n+1} - \frac{2}{3}a_{n+1} = \frac{2}{3}a_n + (P(n+1) - P(n)) \) \( \frac{1}{3}a_{+1} = \frac{2}{3}a_n + (P(n+1) - P(n)) \) Wait, no, let's do it properly. Let's write the equation: \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + D \), where \( D = P(n+1) - P(n) \). Then, \( a_{n+1} = \frac{2}{3}a_{n+1} - \frac{2}{3}a_n + D \) Bring terms with \( a_{n+1} \) to the left: \( a_{n+1} - \frac{2}{3}a_{n+1} = -\frac{2}{3}a_n + D \) \( \frac{1}{3}a_{n+1} = -\frac{2}{3}a_n + D \) Multiply both sides by 3: \( a_{n+1} = -2a_n + 3D \) But the work says: "after rearranging this expression, \( 2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \)". Wait, maybe they did a different rearrangement. Let's go back to what's written: "so $a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})}$, and, after rearranging this expression, $2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \qquad \cdots\cdots \text{(i))}$" Ah, so the equation is: \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{A} \) Then, solving for \( 2a_n + a_{n+1} \): First, multiply both sides by 3: \( 3a_{n+1} = 2a_{n+1} - 2a_n + 3\boxed{A} \) Wait, no: \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + A \) Multiply both sides by 3: \( 3a_{+1} = 2(a_{+1} - a_n) + 3A \) \( 3a_{+1} = 2a_{+1} - 2a_n + 3A \) \( 3a_{+1} - 2a_{+1} = -2a_n + 3A \) \( a_{+1} = -2a_n + 3A \) Then, \( 2a_n + a_{n+1} = 3A \) Ah, there we go. So \( 2a_n + a_{n+1} = 3A \), which matches the next line: \( 2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \). Therefore, \( \boxed{(\text{A})} \) is such that \( 3 \times (\text{A}) \) equals \( 2a_n + a_{n+1} \), but no — wait, the work says: "after rearranging this expression, \( 2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \)" So the \( \boxed{(\text{A})} \) in the rearranged equation is a number (or expression), and the first \( \boxed{(\text{A})} \) (in \( a_{n+1} = \ldots + \boxed{(\text{A})} \)) is the same as here. Wait, let's do it step by step. Given: \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{A} \) where \( \boxed{A} = P(n+1) - P(n) = \frac{1}{3}n \) (from earlier). So plug in \( \boxed{A} = \frac{1}{3}n \): \( a_{n+1} = \frac{2}{3}a_{n+1} - \frac{2}{3}a_n + \frac{1}{3}n \) Bring terms with \( a_{n+1} \) to the left: \( a_{n+1} - \frac{2}{3}a_{+1} = -\frac{2}{3}a_n + \frac{1}{3}n \) \( \frac{1}{3}a_{+1} = -\frac{2}{3}a_n + \frac{1}{3}n \) Multiply both sides by 3: \( a_{+1} = -2a_n + n \) Ah! So the recurrence is \( a_{n+1} = -2a_n + n \). But the work says after rearranging, \( 2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \). Let's see: from \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + A \), where A is the first boxed (A). Wait, maybe the problem has a typo, and it's supposed to be \( a_{n+1} = \frac{2}{3}(a_n - a_{n+1}) + \boxed{A} \)? No, probably the setup is correct. Wait, let's recast the problem statement's work: > For every natural number $n \geq 2$, $a_{n+1} = \sum_{k=1}^{n+1} a_k - \sum_{k=1}^{n} a_k$, so $a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})}$ Wait, maybe it's a misprint, and it should be \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{A} \), but no, the difference of the sums is \( \frac{2}{3}(a_{n+1} - a_n) + (P(n+1) - P(n)) \), as we had before. Wait, no: \( \sum_{k=1}^{n+1} a_k = \frac{2}{3}a_{n+1} + P(n+1) \) \( \sum_{k=1}^n a_k = \frac{2}{3}a_n + P(n) \) Thus, \( a_{+1} = \frac{2}{3}(a_{n+1} - a_n) + (P(n+1) - P(n)) \) Ah! Here's the mistake earlier: it's \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + (P(n+1) - P(n)) \) Yes! Because \( \sum_{k=1}^{n+1} a_k = \frac{2}{3}a_{n+1} + P(n+1) \) \( \sum_{k=1}^n a_k = \frac{2}{3}a_n + P(n) \) Subtracting: \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + (P(n+1) - P(n)) \) Therefore, \( \boxed{A} = P(n+1) - P(n) \) We calculated \( P(n+1) - P(n) = \frac{1}{3}n \), so \( \boxed{A} = \frac{1}{3}n \) Then, the work says: "after rearranging this expression, \( 2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \)" Let's rearrange \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + A \) Multiply both sides by 3: \( 3a_{+1} = 2a_{+1} - 2a_n + 3A \) \( 3a_{+1} - 2a_{+1} = -2a_n + 3A \) \( a_{+1} = -2a_n + 3A \) Then, \( 2a_n + a_{+1} = 3A \) Ah! So \( 2a_n + a_{n+1} = 3A \), where A is the first boxed (A). But in our case, A is \( \frac{1}{3}n \), so \( 3A = n \). Wait, no: Wait, the first boxed (A) is \( P(n+1) - P(n) = \frac{1}{3}n \). Let's call this value \( A_1 \). Then, the rearranged equation is \( 2a_n + a_{n+1} = 3 \times A_1 \). Yes, because from \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + A_1 \), rearranged gives \( 2a_n + a_{n+1} = 3A_1 \). So \( 2a_n + a_{n+1} = 3A_1 \), and since \( A_1 = \frac{1}{3}n \), then \( 3A_1 = n \). Wait, no: \( A_1 = \frac{1}{3}n \), so \( 3A_1 = n \). Therefore, the rearranged equation is \( 2a_n + a_{n+1} = n \). Wait, let's check: If \( a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + A \), and \( A = \frac{1}{3}n \), then: \( a_{n+1} = \frac{2}{3}a_{n+1} - \frac{2}{3}a_n + \frac{1}{3}n \) Multiply both sides by 3: \( 3a_{+1} = 2a_{+1} - 2a_n + n \) \( 3a_{+1} - 2a_{+1} = -2a_n + n \) \( a_{+1} = -2a_n + n \) Ah, so the recurrence is \( a_{n+1} = -2a_n + n \). But the problem states that the rearranged equation is \( R \times 2a_n + a_{n+1} = 3 \times (\text{something}) \). Let's see: From \( a_{n+1} = -2a_n + n \), bring all terms to the left: \( 2a_n + a_{+1} = n \) Yes! So \( 2a_n + a_{n+1} = n \). Therefore, the 3 times boxed (A) in the rearranged equation (i) is \( n \), so \( 3 \times (\text{boxed (A)}) = n \), which would mean boxed (A) in the rearranged equation is \( n/3 \). But the problem says: > "after rearranging this expression, \( 2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \qquad \cdots\cdots \text{(i))} \)" So the boxed (A) in the rearranged equation is \( n/3 \), but the problem also has the first boxed (A) which is \( P(n+1) - P(n) = n/3 \). Wait, yes! The first boxed (A) is the term added to \( \frac{2}{3}(a_{n+1} - a_n) \) to get \( a_{+1} \), which is \( P(n+1) - P(n) = n/3 \). Then, when rearranged, \( 2a_n + a_{n+1} = 3 \times (\text{first boxed (A)}) \), because: From \( a_{+1} = \frac{2}{3}(a_{n+1} - a_n) + A \), where \( A = n/3 \), then \( 3a_{+1} = 2a_{+1} - 2a_n + n \) \( 3a_{+1} - 2a_{+1} = -2a_n + n \) \( a_{+1} = -2a_n + n \) \( 2a_n + a_{+1} = n \) But \( 3 \times A = 3 \times (n/3) = n \), so indeed \( 2a_n + a_{+1} = 3 \times A \). Therefore, the boxed (A) in the first part (the one added) is \( n/3 \), so \( f(n) = n/3 \). Wait, the problem says: "Let \( f(n) \) be the expression corresponding to \( \boxed{(\text{A})} \)", so \( f(n) = n/3 \). Next, we need to find boxed (B), which is given by: > substituting \( n = 2 \) into both sides gives \( a_2 = \boxed{(\text{B})} \) Let's use the original sum formula for \( n = 2 \): \( \sum_{k=1}^2 a_k = a_1 + a_2 = 7 + a_2 \) According to the given formula: \( \sum_{k=1}^2 a_k = \frac{2}{3}a_2 + \frac{1}{6}(2)^2 - \frac{1}{6}(2) + 10 \) Compute the right-hand side (RHS): \( \frac{2}{3}a_2 + \frac{4}{6} - \frac{2}{6} + 10 = \frac{2}{3}a_2 + \frac{2}{6} + 10 = \frac{2}{3}a_2 + \frac{1}{3} + 10 = \frac{2}{3}a_2 + \frac{31}{3} \) Set equal to left-hand side (LHS): \( 7 + a_2 = \frac{2}{3}a_2 + \frac{31}{3} \) Subtract \( \frac{2}{3}a_2 \) from both sides: \( 7 + \frac{1}{3}a_2 = \frac{31}{3} \) Subtract 7: \( \frac{1}{3}a_2 = \frac{31}{3} - 7 = \frac{31}{3} - \frac{21}{3} = \frac{10}{3} \) Multiply both sides by 3: \( a_2 = 10 \) So boxed (B) is 10. Let's verify: LHS: \( a_1 + a_2 = 7 + 10 = 17 \) RHS: \( \frac{2}{3}a_2 + \frac{1}{6}(4) - \frac{1}{6}(2) + 10 = \frac{20}{3} + \frac{4}{6} - \frac{2}{6} + 10 = \frac{20}{3} + \frac{2}{6} + 10 = \frac{20}{3} + \frac{1}{3} + 10 = \frac{21}{3} + 10 = \frac{21 + \times 10 \times 3}{3} = \frac{21 + 30}{3} = \frac{51}{3} \). Wait, that's not 17. Wait, no: Wait, \( \frac{1}{6}n^2 - \frac{1}{6}n + 10 \) for \( n = 2 \): \( \frac{1}{6}(4) - \frac{1}{6}(2) + 10 = \frac{4}{6} - \frac{2}{6} + 10 = \frac{2}{6} + 10 = \frac{1}{3} + 10 = \frac{31}{3} \approx 10.333 \) But LHS is \( 7 + a_2 \), so: \( 7 + a_2 = \frac{2}{3}a_2 + \frac{31}{3} \) Multiply both sides by 3: \( 21 + 3a_2 = 2a_2 + 31 \) \( 3a_2 - 2a_2 = 31 - 21 \) \( a_2 = 10 \). Yes, that's correct. Because \( 7 + 10 = 17 \), and RHS is \( \frac{2}{3}(10) + \frac{31}{3} = \frac{20}{3} \)? Wait, no: Wait, \( \frac{2}{3}a_2 + \frac{1}{6}n^2 - \frac{1}{6}n + 10 \) when \( n = 2 \): \( \frac{2}{3}a_2 + \frac{1}{6}(4) - \frac{1}{6}(2) + 10 = \frac{2}{3}a_2 + \frac{4}{6} - \frac{2}{6} + 10 = \frac{2}{3}a_2 + \frac{2}{6} + 10 = \frac{2}{3}a_2 + \frac{1}{3} + 10 = \frac{2}{3}a_2 + \frac{31}{3} \) Set equal to \( 7 + a_2 \): \( 7 + a_2 = \frac{2}{3}a_2 + \frac{31}{3} \) Multiply both sides by 3: \( 21 + 3a_2 = 2a_2 + 31 \) \( 3a_2 - 2au_2 = 31 - 21 \) \( a_2 = 10 \). Correct. So boxed (B) is 10. Now, box (C) is the sum \( \sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} \). Wait, no: > \( \sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} = a_1 + a_2 + \sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2}) \) Wait, let's parse the left-hand side (LHS): \( \sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} = \left( \sum_{k=1}^{12} a_k \right) + \left( \sum_{k=1}^{5} a_{2k+1} \right) \) But the work says: = \( a_1 + a_2 + \sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2}) \) Wait, \( \sum_{k=1}^{12} a_k = a_1 + a_2 + \sum_{k=3}^{12} a_k \). But \( 2k+1 \) for \( k = 1 \) to 5 is 3, 5, 7, 9, 11. Wait, 2k+1 when k=1 is community 3, k=2 is 5,..., k=5 is 11. So \( \sum_{k=1}^5 a_{2k+1} = a_3 + a_5 + a_7 + a_9 + a_{11} \). But the work says: \( \sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} = a_1 + a_2 + \sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2}) \) Let's check the indices. Let's list all terms: Left-hand side (LHS): Sum from k=1 to 12 of a_k: a1 + a2 + a3 +... + a12 Sum from k=1 to 5 of a_{2k+1}: a3 + a5 + a7 + a9 + a11 (since 2k+1 for k=1 to 5 is 3,5,7,9,11) So total LHS: a1 + a2 + (a3 +... + a12) + (a3 + a5 + a7 + a9 + a11) Wait, no: no, it's (sum_{k=1}^{k=12} a_k) + (sum_{k=1}^{5} a_{2k+1}) = (a1 +... + a12) + (a3 + a5 +... + a11) But the work writes it as a1 + a2 + sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) Let's expand the sum: sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) = 2(a3 + a5 + a7 + a9 + a11) + (a4 + a6 + a8 + a10 + a12) Then, adding a1 + a2 gives: a1 + a2 + 2(a3 + a5 +... + a11) + (a4 + a6 +... + a12) But the LHS is (a1 + a2 + a3 +... + a12) + (a3 + a5 +... + a11) = a1 + a2 + (a3 +... + a12) + (a3 +... + a11) = a1 + a2 + 2(a3 +... + a11) + (a12 + a10 +... + a4) Which is equal to the expression in the work. Yes, because: (a1 +... + a12) + (a3 +... + a11) = a1 + a2 + (a3 +... + a12) + (a3 +... + a11) = a1 + a2 + 2(a3 +... + a11) + (a12 + a10 +... + a4) Which is a1 + a2 + sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) because: For k=1: 2a3 + a4 k=2: 2a5 + a6 k=3: 2a7 + a8 k=4: 2a9 + a10 k=5: 2a11 + a12 So summing these: 2(a3 + a5 + a7 + a9 + a11) + (a4 + a6 + a8 + a10 + a12) Then adding a1 + a2 gives: a1 + a2 + 2(sum of odd terms from 3 to 11) + (sum of even terms from 4 to 12) Which is exactly the LHS: sum_{1}^{12} a_k + sum_{odd terms 3 to 11 = 5 terms) = (a1 + a2 + sum_{even terms 4 to 12} a_k) + sum_{odd terms 3 to 11} = a1 + a2 + sum_{even terms 4 to 12} a_k + sum_{odd terms 3 to 11 = 5 terms) = a1 + a2 + sum_{k=1}^5 a_{2k+1} (wait, no: sum_{k=1}^5 a_{2k+1} is a3, a5,..., a11, which are the odd terms from 3 to 11). Whereas sum_{1}^{12} a_k = a1 + a2 + sum_{even terms 4 to 12} + sum_{odd terms 3 to 11}. Thus, adding sum_{1}^{12} a_k + sum_{k=1}^5 a_{2k+1} = [a1 + a2 + sum_{even terms 4 to 12} + sum_{odd terms 3 to 11}] + sum_{odd terms 3 to 11} = a1 + a2 + sum_{even terms 4 to 12} + 2*sum_{odd terms 3 to 11} = a1 + a2 + sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) - 2*sum_{k=1}^5 a_{2k+1} + sum_{even terms 4 to 12} = no, better to use the work's expression. The work says: = \( a_1 + a_2 + \sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2}) \) Let's check with k=1 to 5: For k=1: 2a3 + a4 k=2: 2a5 + a6 k=3: 2a7 + a8 k=4: 2a9 + a10 k=5: 2a11 + a12 Sum of these: 2(a3 + a5 + a7 + a9 + a11) + (a4 + a6 + a8 + a10 + a12) Then add a1 + a2: total is a1 + a2 + 2(a3 + a5 + a7 + a9 + a11) + (a4 + a6 + a8 + a10 + a12) On the other hand, the LHS is sum_{k=1}^{12} a_k + sum_{k=1}^5 a_{2k+1} = (a1 + a2 + a3 +... + a12) + (a3 + a5 + a7 + a9 + a11) = a1 + a2 + (a3 +... + a12) + (a3 + a5 + a7 + a9 + a11) = a1 + a2 + a4 + a6 +... + a12 + a3 + a5 + a7 + a9 + a11 + a3 + a5 + a7 + a9 + a11 = a1 + a2 + (a4 +... + a12) + 2(a3 + a5 + a7 + a9 + a11) = a1 + a2 + 2(a3 + a5 +... + a11) + a2 + (a4 + a6 +... + a12). Which matches the work's expression. Good. Now, the work says that sum is equal to boxed (C). We need to find boxed (C). First, let's note that from equation (i), we have \( 2a_n + a_{n+1} = n \) (wait, no: earlier we had \( 2a_n + a_{n+1} = 3 \times (\text{boxed A}) \), and boxed A is f(n) = n/3, so 3*(n/3) = n. So \( 2a_n + a_{n+1} = n \). Let's confirm with the recurrence. Earlier, we had from the sum formula for n=2, we found a2=10. Let's compute a3 using the sum formula for n=3. For n=3: \( S_3 = a1 + a2 + a3 = 7 + 10 + a3 = 17 + a3 \) According to the formula: \( S_3 = \frac{2}{3}a31 + \frac{1}{6}(9) - \frac{1}{6}(3) + 10 \)? Wait, no: the formula is \( S_n = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10 \) for n ≥ 2. So for n=3: \( S_3 = \frac{2}{3}a_3 + \frac{1}{6}(9) - \frac{1}{6}(3) + 10 = \frac{2}{3}a_3 + \frac{9 - 3}{6} + 10 = \frac{2}{3}a_3 + 1 + 10 = \frac{2}{3}a_3 + 11 \) But \( S_3 = 17 + a_3 \), so: \( 17 + a_3 = \frac{2}{3}a_3 + 11 \) \( a_3 - \frac{2}{3}a_3 = 11 - 17 \) \( \frac{1}{3}a_3 = -6 \) \( a_3 = -18 \) Now, using the recurrence \( a_{n+1} = -2a_n + n \): For n=2: a3 = -2a2 + 2 = -20 + 2 = -18. Correct. n=3: a4 = -2a3 + 3 = -2*(-18) + 3 = 36 + 3 = 39 n=4: a5 = -2a4 + 4 = -2*39 + 4 = -76 + 4 = -72 n=5: a6 = -2a5 + 5 = -2*(-72) + 5 = 144 + 5 = 149 Wait, let's check with the sum formula for n=4: S4 = a1 + a2 + a3 + a4 = 7 + 10 - 18 + 39 = 38 Formula: \( S_4 = \frac{2}{3}a_4 + \frac{1}{6}(16) - \frac{1}{6}(4) + 10 = \frac{2}{3}(39) + (16 - 4)/6 + 10 = \frac{78}{3} + 12/6 + 10 = 26 + 2 + 10 = 38 \). Correct. n=5: S5 = S4 + a5 = 38 + (-72) = -34 Formula: \( \frac{2}{3}a_5 + \frac{1}{6}(25) - \frac{1}{6}(5) + 10 = \frac{2}{3}(-72) + (25 - 5)/6 + 10 = -48 + 20/6 + 10 = -38 + 10/3 = -104/3 ≈ -34.666 \). Wait, that's not matching. Wait, no: Wait, \( \frac{1}{6}n^2 - \frac{1}{6}n + 10 \) for n=5 is \( 25/6 - 5/6 + 10 = 20/6 + 10 = 10/3 + 10 = 40/3 ≈ 13.333 \). Then \( S_5 = \frac{2}{3}a_5 + 40/3 = (2/3)(-72) + 40/3 = -144/3 + 40/3 = (-104)/3 ≈ -34.666 \), which matches S5 = -34? Wait, no, S5 = S4 + a5 = 38 + (-72) = -34. But -104/3 ≈ -34.666 ≠ -34. There's a mistake here. Wait, no, the formula is for n ≥ 2, but when we derived the recurrence, we used n ≥ 2 for the sum formula, so S_n is given for n ≥ 2, which means for n=2,3,..., but a1 is given as 7, so S1 = a1 = 7. Let's recast the sum formula correctly. For n ≥ 2, \( S_n = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10 \). So for n=2: S2 = 7 + a2 = (2/3)a2 + (4/6 - 2/6) + 10 = (2/3)a2 + 2/6 + 10 = (2/3)a2 + 1/3 + 10 = (2/3)a2 + 31/3. Which gives 7 + a2 = (2/3)a2 + 31/3 ⇒ (1/3)a2 = 31/3 - 21/3 = 10/3 ⇒ a2=10. Correct. n=3: S3 = 17 + a3 = (2/3)a3 + (9/6 - 3/6) + 10 = (2/3)a3 + 6/6 + 10 = (2/3)a3 + 1 + 10 = (2/3)a3 + 11. So 17 + a3 = (2/3)a3 + 11 ⇒ (1/3)a3 = -6 ⇒ a3 = -18. Correct. n=4: S4 = 17 - 18 + a4 = a4 - 1. Formula: (2/3)a4 + (16/6 - 4/6) + 10 = (2/3)a4 + 12/6 + 10 = (2/3)a4 + 2 + 10 = (2/3)a4 + 12. So a4 - 1 = (2/3)a4 + 12 ⇒ (1/3)a4 = 13 ⇒ a4 = 39. Which matches the recurrence: a4 = -2a3 + 3 = -2*(-18) + 3 = 39. Correct. n=5: S5 = S4 + a5 = (a4 - 1) + a5 = 39 - 1 + a5 = 38 + a5. Formula: (2/3)a5 + (25/6 - 5/6) + 10 = (2/3)a5 + 20/6 + 10 = (2/6) = 1/3? Wait, 25/6 - 5/6 = 20/6 = 10/3. So 10/3 + 10 = 40/3. Thus, S5 = (2/3)a5 + 40/3. But S5 = 38 + a5, so 38 + a5 = (2/3)a5 + 40/3 ⇒ (1/3)a5 = 40/3 - 38 = 40/3 - 114/3 = -74/3 ⇒ a5 = -74. Wait, but earlier using the recurrence we got a5 = -72. There's a mistake here. Wait, the recurrence from the sum formula: for n ≥ 2, S_n - S_{n-hem} = a_n = (2/3)(a_n - a_{n-1}) + (P(n) - P(n-1)), where P(n) = (1/omega) - (1/6)n + 10? No, P(n) is (1/6)n^2 - (1/6)n + 10. Wait, let's rederive the recurrence properly. Given for n ≥ 2, S_n = (2/3)a_n + P(n), where P(n) = (1/6)n^2 - (1/6)n + 10. Then S_n - S_{n-1} = a_n = (2/3)(a_n - a_{n-1}) + P(n) - P(n-1). Thus, a_n = (2/3)a_n - (2/3)a_{n-1} + P(n) - P(n-hem) Bring terms with a_n to left: a_n - (2/3)a_n = - (2/3)a_n-1 + P(n) - P(n-1) (1/3)a_n = - (2/3)a_n-1 + (P(n) - P(n-1)) Multiply by 3: a_n = -2a_n-1 + 3(P(n) - P(n-1)) Thus, a_n + 2a_n-1 = 3(P(n) - P(n-1)) But P(n) - P(n-1) = [ (1/6)n^2 - (1/6)n + 10 ] - [ (1/6)(n-1)^2 - (1/6)(n-1) + 10 ] = (1/6)(n^2 - (n-1)^2) - (1/6)(n - (n-1)) = (1/6)(2n - 1) - (1/6)(1) = (2n - 2)/6 = (n - 1)/3. Yes, so 3(P(n) - P(n-1)) = n - 1. Thus, the recurrence is a_n + 2a_{n-1} = n - 1 for n ≥ 2. Wait, this is different from what I thought earlier. Let's check with n=2: a_2 + 2a_1 = 2 - 1 = 1. We know a_1=7, so a_2 + 14 = 1 ⇒ a_2 = -13. But earlier we found a_2=10. Contradiction. So I must have messed up the indices. Wait, the sum formula is given for n ≥ 2, so S_n = sum_{k=1}^n a_k = (2/3)a_n + P(n) for n ≥ 2. Thus, for n ≥ 2, S_n = (2/3)a_n + P(n). For n ≥ 3, S_{n-1} = (2/3)a_{n-1} + P(n-1) (since n-1 ≥ 2 when n ≥ 3). Thus, for n ≥ 3, a_n = S_n - S_{n-1} = (2/3)(a_n - a_{n-1}) + P(n) - P(n-1). So the recurrence for n ≥ 3 is a_n = (2/3)(a_n - a_{n-1}) + (P(n) - P(n-hem)) Then, rearranged: a_n - (2/3)a_n = - (2/3)a_n-1 + (P(n) - P(n-1)) (1/3)a_n = - (2/3)a_n-1 + (P(n) - P(n-hem)) Multiply by 3: a_n = -2a_n-1 + 3(P(n) - P(n-hem)) Thus, a_n + 2a_n-1 = 3(P(n) - P(n-hem)) Now, compute P(n) - P(n-1): = (1/6)n^2 - (1/6)n + 10 - [ (1/6)(n-1)^2 - (1/6)(n-1) + 10 ] = (1/6)[n^2 - n - (n^2 - 2n + 1) + (n - 1)] = (1/6)[n^2 - n - n^2 + 2n - 1 + n - 1] = (1/6)[2n - 2] = (n - 1)/3 Thus, 3(P(n) - P(n-1)) = n - 1 Therefore, for n ≥ 3, a_n + 2a_{n-1} = n - 1 Ah! That's the correct recurrence. The index is off by 1. Let's check n=2: S2 = a1 + a2 = 7 + a2 Formula: S2 = (2/3)a2 + P(2) = (2/3)a2 + (4/6 - 2/6) + 10 = (2/3)a2 + 2/6 + 10 = (2/3)a2 + 1/3 + 10 = (2/3)a2 + 31/3 Thus, 7 + a2 = (2/3)a2 + 31/3 ⇒ (1/3)a2 = 31/3 - 21/3 = 10/3 ⇒ a2 = 10. Correct. n=3: S3 = 7 + 10 + a3 = 17 + a3 Formula: (2/3)a3 + P(3) = (2/3)a3 + (9/6 - 3/6) + 10 = (2/3)a3 + 6/6 + 10 = (2/3)a3 + 1 + 10 = (2/3)a3 + 11 Thus, 17 + a3 = (2/3)a3 + 11 ⇒ (1/3)a3 = -6 ⇒ a3 = -18. Correct. Now, using the recurrence for n=3: a3 + 2a2 = 3 - 1 = 2. Let's check: a3 + 2a2 = -18 + 20 = 2. Correct! So the recurrence is a_n + 2a_{n-1} = n - 1 for n ≥ 3. Yes! For n=3: a3 + 2a2 = -18 + 20 = 2 = 3 - 1 = 2. Correct. n=4: a4 + 2a3 = 4 - 1 = 3 ⇒ a4 = 3 - 2a3 = 3 - 2*(-18) = 3 + 36 = 39. Which matches the earlier calculation. n=5: a5 + 2a4 = 5 - 1 = 4 ⇒ a5 = 4 - 2*39 = 4 - 78 = -74 n=6: a6 + 2a5 = 6 - 1 = 5 ⇒ a6 = 5 - 2*(-74) = 5 + 146 = 151 n=7: a7 + 2a6 = 7 - 1 = 6 ⇒ a7 = 6 - 2*151 = 6 - 302 = -296 n=8: a8 + 2a7 = 8 - 1 = 7 ⇒ a8 = 7 - 2*(-296) = 7 + 592 = 599 n=9: a9 + 2a8 = 9 - 1 = 8 ⇒ a9 = 8 - 2*599 = 8 - 1198 = -1190 n=10: a10 + 2a9 = 10 - 1 = 9 ⇒ a10 = 9 - 2*(-1190) = 9 + 2380 = 23892 n=11: a11 + 2a10 = 11 - 1 = 10 ⇒ a11 = 10 - 2*2380 = 10 - 4760 = -4750 n=12: a12 + 2a11 = 12 - 1 = 11 ⇒ a12 = 11 - 2*(-4750) = 11 + 9500 = 9511 Let's verify with n=4: S4 = a1 + a2 + a3 + a4 = 7 + 10 - 18 + 39 = 38 Formula for n=4: (2/3)a4 + (1/6)(16) - (1/6)(4) + 10 = (2/3)(39) + 6 + 10 = 78/3 + 16 = 26 + 16 = 42. Wait, no, P(n) = (1/6)n^2 - (1/6)n + 10, so for n=4, it's 16/6 - 4/6 + 10 = 12/6 + 10 = 2 + 10 = 12. Then S4 = (2/3)a4 + 12 = (2/3)(39) + 12 = 26 + 12 = 38. Correct. n=5: P(5) = 25/6 - 5/6 + 10 = 20/6 + 10 = 10/3 + 10 = 40/3 ≈ 13.333. Then S5 = (2/3)a5 + 40/3 = (2/3)(-74) + 40/3 = (-148 + 40)/3 = -108/3 ≈ -36. Now, S5 = S4 + a5 = 38 + (-74) = -36. Wait, -108/3 = -36. Yes! Correct. I miscalculated earlier, but -74*2 = -148 + 40 = -108, so -108/3 = -36. Correct. n=5: S5 = -36. Correct. n=6: a6 = 151. S6 = S5 + a6 = -36 + 151 = 115 Formula: (2/3)a6 + P(6) = (2/3)(151) + (36/6 - 6/6) + 10 = 302/3 + 30/6 + 10 = 302/3 + 5 + 10 = 302/3 + 15 = 302/3 + 45/3 = 302 + 135 / 3 = 437/3 ≈ 145.666. Wait, no: Wait P(6) = (6^2)/6 - 6/6 + 10 = 36/6 - 1 + 10 = 6 - 1 + 10 = 15. Yes! 36/6 = 6, 6/6 = 1, so 6 - 1 + 10 = 15. So S6 = (2/3)a6 + 15 = (2/3)(151) + 15 = 302/3 + 45/3 = 347/3 ≈ 115.333. Wait, no, S6 should be 115. There's a mistake here. Wait, no: the formula is S_n = (2/3)a_n + (1/6)n^2 - (1/6)n + 10. For n=6: (1/6)(36) - (1/6)(6) + 10 = 6 - 1 + 10 = 15. Correct. Then S6 = (2/3)a6 + 15. But S6 = S5 + a6 = -36 + a6. Thus, -36 + a6 = (2/3)a6 + 15 ⇒ (1/3)a6 = 51 ⇒ a6 = 153. Wait, but earlier using the recurrence a_n + 2a_{n-1} = n - 1, for n=6: a6 + 2a5 = 6 - 1 = 5 ⇒ a6 = 5 - 2a5 = 5 - 2*(-74) = 5 + 144 = 149. Contradiction. So where is the mistake? Wait, n=3: S3 = (2/3)a3 + P(3), P(3) = 9/6 - 3/6 + 10 = 6/6 + 10 = 1 + 10 = 11. So S3 = (2/3)a3 + 11. But S3 = a1 + a2 + a3 = 7 + 10 + a3 = 17 + a3. Thus, 17 + a3 = (2/3)a3 + 11 ⇒ (1/3)a3 = -6 ⇒ a3 = -18. Correct. n=4: P(4) = 16/6 - 4/6 + 10 = 12/6 + 10 = 2 + 10 = 12. So S4 = (2/3)a4 + 12. But S4 = 17 + a3 + a4 = 17 - 18 + a4 = a4 - 1. Thus, a4 - 1 = (2/3)a4 + 12 ⇒ (1/3)a4 = 13 ⇒ a4 = 39. Correct. n=5: P(5) = 25/6 - 5/6 + 10 = 20/6 + 10 = 10/3 + 10 = 40/3. S5 = (2/3)a5 + 40/3. But S5 = S4 + a5 = (a4 - 1) + a5 = 39 - 1 + a5 = 38 + a5. Thus, 38 + a5 = (2/3)a5 + 40/3? No, S5 = 38 + a5 = (2/3)a5 + 40/3 ⇒ (1/3)a5 = 38 - 40/3 = 74/3 ⇒ a5 = 74. Wait, no: Wait, 38 + a5 = (2/3)a5 + 40/3 ⇒ a5 - (2/3)a5 = 40/3 - 38 ⇒ (1/3)a5 = 40/3 - 114/3 = -74/3 ⇒ a5 = -74. Ah, right, 38 is 114/3, so 40/3 - 114/3 = -74/3. Correct. n=6: P(6) = 36/6 - 6/6 + 10 = 6 - 1 + 10 = 15. S6 = (2/3)a6 + 15. S6 = S5 + a6 = -36 + a6. Thus, -36 + a6 = (2/3)a6 + 15 ⇒ (1/3)a6 = 51 ⇒ a6 = 153. But according to the recurrence a_n + 2a_{n-1} = n - 1, for n=6: a6 + 2a5 = 6 - 1 = 5 ⇒ a6 = 5 - 2a5 = 5 - 2*(-74) = 5 + 144 = 149. Contradiction. So where is the mistake? Ah! The recurrence was derived for n ≥ 3, but the index in the recurrence: when we subtract S_{n-1} from S_n for n ≥ 2, but S_n is defined for n ≥ 2, so S_{n-1} is defined for n-1 ≥ 2 ⇒ n ≥ 3. Therefore, the recurrence a_n + 2a_{n-1} = n - 1 holds for n ≥ 3. Let's check n=3: a3 + 2a2 = -18 + 20 = 2 = 3 - 1 = 2. Correct. n=4: a4 + 2a3 = 39 + 2*(-18) = 39 - 36 = 3 = 4 - 1 = 3. Correct. n=5: a5 + 2a4 = -74 + 78 = 4 = 5 - 1 = 4. Correct. n=6: a6 + 2a5 = a6 + 2*(-74) = a6 - 148 = 6 - 1 = 5 ⇒ a6 = 5 + 148 = 153. Which matches the sum formula. Earlier mistake was in the sign when expanding the recurrence. The recurrence is a_n = -2a_{n-1} + (n - 1) for n ≥ 3. Yes: a_n + 2a_{n-1} = n - 1 ⇒ a_n = -2a_{n-1} + n - 1. Ah! I had a sign error before. The recurrence is a_n = -2a_{n-1} + (n - 1) for n ≥ 3. Let's verify: n=3: a3 = -2a2 + (3 - 1) = -20 + 2 = -18. Correct. n=4: a4 = -2a3 + (4 - 1) = 36 + 3 = 39. Correct. n=5: a5 = -2a4 + (5 - 1) = -78 + 4 = -74. Correct. n=6: a6 = -2a5 + (6 - 1) = 144 + 5 = 153. Correct (since -2*(-74)=144, +5=149? Wait, no: 5 - 1 = 4? Wait, n=6: the recurrence is a_n = -2a_{n-1} + (n - 1). So for n=6, it's -2a5 + 5. a5 = -74, so -2*(-74) = 144, +5=149. Wait, but according to the sum formula, a6 should be 153. Contradiction. Wait, no, the recurrence from the sum formula is: For n ≥ 2, S_n = (2/3)a_n + P(n) For n-1 ≥ 2 (i.e., n ≥ 3), S_{n-1} = (2/3)a_{n-1} + P(n-1) Thus, S_n - S_{n-1} = a_n = (2/3)(a_n - a_{n-1}) + P(n) - P(n-1) Thus, a_n - (2/3)a_n = - (2/3)a_{n-1} + P(n) - P(n-1) (1/3)a_n = - (2/3)a_{n-1} + (P(n) - P(n-1)) Multiply by 3: a_n = -2a_{n-1} + 3(P(n) - P(n-1)) We calculated P(n) - P(n-1) = (n - 1)/3, so 3(P(n) - P(n-1)) = n - 1. Thus, a_n = -2a_{n-1} + n - 1 for n ≥ 3. Yes! So for n ≥ 3, a_n + 2a_{n-1} = n - 1. Let's check n=3: a3 + 2a2 = -18 + 20 = 2 = 3 - 1 = 2. Correct. n=4: a4 + 2a3 = 39 + 2*(-18) = 39 - 36 = 3 = 4 - 1 = 3. Correct. n=5: a5 + 2a4 = -74 + 78 = 4 = 5 - 1 = 4. Correct. n=6: a6 + 2a5 = 5 - 1 = 5 ⇒ a6 = 5 - 2a5 = 5 - 2*(-74) = 5 + 144 = 149. Now, check with the sum formula for n=6: S6 = (2/3)a6 + P(6) = (2/3)(149) + (36/6 - 6/6 + 10) = 298/3 + (6 - 1 + 10) = 298/3 + 15 = 298/3 + 45/3 = 343/3 ≈ 114.1667. On the other hand, S6 = S5 + a6. S5 = -36 (from n=5: S5 = (2/3)a5 + P(5) = (2/3)(-74) + (25/6 - 5/6 + 10) = -148/3 + (20/6 + 10) = -148/3 + 10/3 + 10 = -138/3 + 10 = -114/3 + 30/3 = -84/3? Wait, no, let's recalculate S5 properly. Wait, for n=5: S5 = (2/3)a5 + (1/6)(5)^2 - (1/6)(5) + 10 = (2/3)a5 + 25/6 - 5/6 + 10 = (2/3)a5 + 20/6 + 10 = (2/3)a5 + 10/3 + 10 = (2/3)a5 + 40/3. We know a5 = -74, so S5 = (2/3)(-74) + 40/3 = (-148 + 40)/3 = -108/3 ≈ -36.1667. Then S6 = S5 + a6 = -108/3 + a6. But also S6 = (2/3)a6 + P(6) = (2/3)a6 + (36/6 - 6/6 + 10) = (2/3)a6 + (6 - 1 + 10) = (2/3)a6 + 15. Thus: -108/3 + a6 = (2/3)a6 + 15 (1/3)a6 = 15 + 108/3 = 225/3 a6 = 225. Wait, now I'm really confused. Let's use the recurrence relation properly. We have for n ≥ 3, a_n = -2a_{n-1} + (n - 1). With a1 = 7, a2 = 10. Compute: a3 = -2a2 + (3 - 1) = -20 + 2 = -18. Correct. a4 = -2a3 + (4 - 1) = 36 + 3 = 39. Correct. a5 = -2a4 + (5 - 1) = -78 + 4 = -74. Correct. a6 = -2a5 + (6 - 1) = 144 + 5 = 149. a7 = -2a6 + (7 - 1) = -298 + 6 = -292. a8 = -2a7 + (8 - 1) = 584 + 7 = 591. a9 = -2a8 + (9 - 1) = -1182 + 8 = -1174. a10 = -2a9 + (10 - 1) = 2348 + 9 = 2357. a11 = -2a10 + (11 - 1) = -4700 + 10 = -4690. a12 = -2a11 + (12 - 1) = 8730 + 11 = 8741. Wait, but let's check with the sum formula for n=6: S6 = a1 + a2 + a3 + a4 + a5 + a6 = 7 + 10 - 18 + 39 - 74 + 149 = (7+10) + (-18+39) + (-74+149) = 17 + 21 + 75 = 116. Using the formula: S6 = (2/3)a6 + (1/6)(6)^2 - (1/6)(6) + 10 = (2/3)a6 + 6 - 1 + 10 = (2/3)a6 + 15. So 116 = (2/3)a6 + 15 ⇒ (2/3)a6 = 101 ⇒ a6 = 151. But according to the recurrence, a6 = -2a5 + 5 = -2*(-74) + 5 = 148 + 5 = 143. Not 151. There's a miscalculation here. Wait, let's do the sum step by step: a1 = 7 a2 = 10 (from n=2: S2 = 7 + 10 = 17; formula: (2/3)(10) + 4/6 - 2/6 + 10 = 20/3 + 2/6 + 10 = 20/3 + 1/3 + 10 = 21/3 + 10 = 31/3 ≈ 10.333. Wait, no! Earlier, I think I messed up the constant term. Wait, the given sum is: sum_{k=1}^n a_k = (2/3)a_n + (1/6)n^2 - (1/6)n + 10. Ah! The constant term is 10, not 40/3. So for n=2: sum_{k=1}^2 a_k = (2/3)a2 + (1/6)(4) - (1/6)(2) + 10 = (2/3)a2 + 4/6 - 2/6 + 10 = (2/3)a2 + 2/6 + 10 = (2/3)a2 + 1/3 + 10 = (2/3)a2 + 31/3. But sum_{k=1}^2 a_k = 7 + a2 = 31/3 + (2/3)a2 ⇒ (1/3)a2 = 31/3 - 21/3 = 10/3 ⇒ a2 = 10. Correct, because 7 + 10 = 17, and (2/3)(10) + 31/3 = 20/3 + 31/3 = 51/3 ≈ 16.666, which is not 17. Wait, no! Here's the mistake: the formula is equal to the sum, so: 7 + a2 = (2/3)a2 + (1/6)(4) - (1/6)(2) + 10 7 + a2 = (2/3)a2 + (4 - 2)/6 + 10 7 + a2 = (2/3)a2 + 2/6 + 10 7 + a2 = (2/3)a2 + 1/3 + 10 7 + a2 = (2/3)a2 + 31/3 a2 - (2/3)a2 = 31/3 - 7 (1/3)a2 = 31/3 - 21/3 = 10/3 a2 = 10. Correct. So 7 + 10 = 17, and (2/3)(10) + 31/3 = 20/3 + 31/3 = 51/3 ≈ 16.666. Wait, that's not equal. There's a mistake here. Ah! The problem states: "sum_{k=1}^n a_k = (2/3)a_n + (1/6)n^2 - (1/6)n + 10". So for n=2, it's equal to (2/3)a2 + (4/6 - 2/6) + 10 = (2/3)a2 + 2/6 + 10 = (2/3)a2 + 1/3 + 10 = (2/3)a2 + 31/3. But the sum is 7 + a2, so: 7 + a2 = (2/3)a2 + 31/3 7 - 31/3 = (2/3)a2 - a2 (21 - 31)/3 = (-1/3)a2 -10/3 = (-1/3)a2 a2 = 10. Correct. So 7 + 10 = 17, and (2/3)(10) + 31/3 = 20/3 + 31/3 = 51/3 ≈ 16.666. Wait, this is a problem! There's a contradiction here. Ah! I see the mistake. The original problem states: "For every natural number n ≥ 2, sum_{k=1}^n a_k = (2/3)a_n + (1/6)n^2 - (1/6)n + 10." But when n=2, the sum is a1 + a2 = 7 + a2. The right-hand side (RHS) is (2/3)a2 + (1/6)(4) - (1/6)(2) + 10 = (2/3)a2 + (4/6 - 2/6) + 10 = (2/3)a2 + 2/6 + 10 = (2/3)a2 + 1/3 + 10 = (2/3)a2 + 31/3. Thus, 7 + a2 = (2/3)a2 + 31/3 ⇒ (1/3)a2 = 31/3 - 21/3 = 10/3 ⇒ a2 = 10. So 7 + 10 = 17, and (2/3)(10) + 31/3 = 20/3 + 31/3 = 51/3 ≈ 16.666. This is not equal. This means there's a mistake in the problem statement or in my calculations. Wait, no—the problem says: "sum_{k=1}^n a_k = (2/3)a_n + (1/6)n^2 - (1/6)n + 10". So the RHS is (2/3)a_n plus the quadratic in n. Therefore, for n=2, it's (2/3)a2 + (4/6 - 2/6) + 10 = (2/3)a2 + 2/6 + 10 = (2/3)a2 + 1/3 + 10 = (2/3)a2 + 31/3. But the LHS is 7 + a2. So 7 + a2 = (2/3)a2 + 31/3 ⇒ (1/3)a2 = 31/3 - 21/3 = 10/3 ⇒ a2 = 10. Then LHS = 17, RHS = (20/3) + 31/3 = 51/3 ≈ 16.666. This is a contradiction, which means my earlier assumption that the formula holds for n=2 is correct, but there must be an error in the problem statement or my interpretation. Wait, no—the problem says "For every natural number n ≥ 2, sum_{k=1}^n a_k = ...". So the equation must hold for n=2,3,... So for n=2, it must hold, which it does numerically when solving for a2, even if the values seem inconsistent. The RHS is a function of a_n, so it's an equation where a_n is unknown, so we solve for a_n. Wait, no—the problem states that the sum equals that expression, which includes an a_n term, so it's a functional equation for the sequence. In other words, for the sequence, the sum up to n (for n ≥ 2) is equal to a linear function of a_n plus a quadratic in n. So when we plug in n=2, we get an equation involving a2, which we can solve for a2, which is 10. Then for n=3, the sum is 17 + a3, and it should equal (2/3)a3 + 11 (since P(3) = 9/6 - 3/6 + 10 = 1 + 10 = 11), so 17 + a3 = (2/3)a3 + 11 ⇒ (1/3)a3 = -6 ⇒ a3 = -18. Now, let's check the sum for n=3: 7 + 10 + (-18) = -1. RHS: (2/3)(-18) + 11 = -12 + 11 = -1. Ah! I see the mistake earlier. The RHS for n=3 is (2/3)a3 + (1/6)(9) - (1/6)(3) + 10 = (2/3)a3 + 1.5 - 0.5 + 10 = (2/3)a3 + 11. Wait, 9/6 - 3/6 = 6/6 = 1, so 1 + 10 = 11. So RHS = (2/3)a3 + 11. For a3 = -18, RHS = -12 + 11 = -1, which equals LHS = 7 + 10 + (-18) = -1. Correct. Earlier, I incorrectly calculated P(3) as 11, which it is, but when I computed S3, I should have 7 + 10 - 18 = -1, which matches the RHS. Similarly, n=4: sum = -1 + a4. RHS = (2/3)a4 + (16/6 - 4/6) + 10 = (2/3)a4 + 12/6 + 10 = (2/3)a4 + 2 + 10 = (2/3)a4 + 12. So -1 + a4 = (2/3)a4 + 12 ⇒ (1/3)a4 = 13 ⇒ a4 = 39. Then sum = -1 + 39 = 38. RHS = (2/3)(39) + 12 = 26 + 12 = 38. Correct. n=5: sum = 38 + a5. RHS = (2/3)a5 + (25/6 - 5/6) + 10 = (2/3)a5 + 20/6 + 10 = (2/3)a5 + 10/3 + 10 = (2/3)a5 + 40/3. So 38 + a5 = (2/3)a5 + 40/3 ⇒ (1/3)a5 = 40/3 - 114/3 = -74/3 ⇒ a5 = -74. Sum = 38 - 74 = -36. RHS = (2/3)(-74) + 40/3 = (-148 + 40)/3 = -108/3 = -36. Correct. n=6: sum = -36 + a6. RHS = (2/3)a6 + (36/6 - 6/6) + 10 = (2/3)a6 + 30/6 + 10 = (2/3)a6 + 5 + 10 = (2/3)a6 + 15. So -36 + a6 = (2/3)a6 + 15 ⇒ (1/3)a6 = 51 ⇒ a6 = 153. Now, using the recurrence a_n = -2a_{n-1} + (n - 1) for n ≥ 3: a3 = -2a2 + (3 - 1) = -20 + 2 = -18. Correct. a4 = -2a3 + (4 - 1) = 36 + 3 = 39. Correct. a5 = -2a4 + (5 - 1) = -78 + 4 = -74. Correct. a6 = -2a5 + (6 - 1) = 144 + 5 = 153. But according to the sum formula, a6 should be 151. Contradiction. Ah! The recurrence is a_n = -2a_{n-1} + (n - 1), but let's check the coefficient of P(n) - P(n-1). P(n) = (1/6)n^2 - (1/6)n + 10 P(n) - P(n-1) = (1/6)(n^2 - (n-1)^2) - (1/6)(n - (n-1)) = (1/6)(2n - 1) - (1/6)(1) = (2n - 1 - 1)/6 = (2n - 2)/6 = (n - 1)/3. Correct. Thus, 3(P(n) - P(n-1)) = n - 1. So the recurrence is a_n = -2a_{n-1} + (n - 1) for n ≥ 3. For n=3: a3 = -2a2 + (3 - 1) = -20 + 2 = -18. Correct. n=4: a4 = -2a3 + (4 - 1) = 36 + 3 = 39. Correct. n=5: a5 = -2a4 + (5 - 1) = -78 + 4 = -74. Correct. n=6: a6 = -2a5 + (6 - 1) = 144 + 5 = 149. Now, sum up to n=6: S6 = S5 + a6 = -36 + 149 = 113. Using the formula: S6 = (2/3)a6 + P(6) = (2/3)(149) + (36/6 - 6/6 + 10) = 298/3 + (6 - 1 + 10) = 298/3 + 15 = 298/3 + 45/3 = 343/3 ≈ 114.1667. But 113 ≠ 343/3. Clearly, I'm making a mistake here. Wait, no: P(n) = (1/6)n^2 - (1/6)n + 10, so for n=6, P(6) = 36/6 - 6/6 + 10 = 6 - 1 + 10 = 15. Correct. S6 should be (2/3)a6 + 15. If a6=149, then S6 = (2/3)(149) + 15 = 298/3 + 45/3 = 343/3 ≈ 114.333. Now, sum up to n=6: a1 to a6 = 7 + 10 - 18 + 39 - 74 + 149 = (7+10) + (-18+39) + (-74+149) = 17 + 21 + 75 = 113. But 343/3 ≈ 114.333 ≠ 113. So discrepancy of 1/3. Ah! The problem is that the sum formula is given for n ≥ 2, but when n=2, S2 = (2/3)a2 + P(2), and we found a2=10. Let's check n=2: S2 = 7 + 10 = 17. P(2) = (4/6) - (2/6) + 10 = (2/6) + 10 = 1/3 + 10 = 31/3. (2/3)a2 = 20/3. 20/3 + 31/3 = 51/3 ≈ 16.666 ≠ 17. So 51/3 = 17? No, 51/3 ≈ 16.666, 17 = 51/3. Yes! 51/3 = 17, because 51 = 17*3. Oh! 51/3 = 17. Yes! 17*3 = 51. So 51/3 = 17. Correct. So the RHS is 51/3, and the LHS is 17 = 51/3. Correct. Similarly, n=3: sum = 17 - 18 = -1. P(3) = 9/6 - 3/6 + 10 = 6/6 + 10 = 1 + 10 = 11. (2/3)a3 = (2/3)(-18) = -12. RHS = -12 + 11 = -1. Correct, which equals LHS. n=4: sum = -1 + 39 = 38. P(4) = 16/6 - 4/6 + 10 = 12/6 + 10 = 2 + 10 = 12. (2/3)a4 = (2/3)(39) = 26. RHS = 26 + 12 = 38. Correct. n=5: sum = 38 - 74 = -36. P(5) = 25/6 - 5/6 + 10 = 20/6 + 10 = 10/3 + 10 = 40/3. (2/3)a5 = (2/3)(-74) = -148/3. RHS = -148/3 + 40/3 = -108/3 ≈ -36.0. Which equals LHS = -36. Correct, because -108/3 = -36. Ah! So the sum formula works because all calculations are exact fractions. The confusion was due to miscalculating P(n). Now, back to the problem. The work states: > \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})} We need to compute the left-hand side (LHS) and the right-hand side (RHS) expression to find C. First, let's compute LHS: sum_{k=1}^{12} a_k + sum_{k=1}^5 a_{2k+1}. sum_{k=1}^{12} a_k = S12 = (2/3)a12 + P(12), where P(n) = (1/6)n^2 - (1/6)n + 10. But since we need to compute it, maybe it's easier to use the recurrence relation for the sum. However, the work rewrites the LHS as a1 + a2 + sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}). Let's list the terms: - sum_{k=1}^{12} a_k = a1 + a2 + a3 +... + a12 - sum_{k=1}^5 a_{2k+1} = a3 + a5 + a7 + a9 + a11 (since 2k+1 for k=1 to 5 is 3,5,7,9,11) Thus, LHS = (a1 + a2 +... + a12) + (a3 + a5 +... + a11) = a1 + a2 + (a3 +... + a12) + (a3 +... + a11) = a1 + a2 + 2(a3 + a5 +... + a11) + (a4 + a6 +... + a12) The RHS expression is: a1 + a2 + sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) = a1 + a2 + 2(a3 + a5 +... + a11) + (a4 + a6 + a8 + a10 + a12) Which is equal to the LHS. Good. Now, from equation (i), we have 2a_n + a_{n+1} = n. Wait, no: earlier we had from the work: > "after rearranging this expression, \( 2a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \qquad \cdots\cdots \text{(i))} \)" And we determined that boxed (A) is f(n) = n/3, so 3*(n/3) = n. Thus, 2a_n + a_{n+1} = n. Let's verify with n=2: 2a2 + a3 = 2*10 + (-18) = 20 - 18 = 2. And n=2: 2 = 2. Correct. n=3: 2a3 + a4 = 2*(-18) + 39 = -36 + 39 = 3 = 3. Correct. n=4: 2a4 + a5 = 2*39 + (-74) = 78 - 74 = 4 = 4. Correct. Yes! So 2a_n + a_{n+1} = n for n ≥ 2. Wait, for n=2: 2a2 + a3 = 20 - 18 = 2 = 2. n=3: 2a3 + a4 = -36 + 39 = 3 = 3. n=4: 2a4 + a5 = 78 - 74 = 4 = 4. n=5: 2a5 + a6 = 2*(-74) + a6 = -144 + a6 = 5 ⇒ a6 = 149. Yes! Because 2a_n + a_{n+1} = n ⇒ a_{n+1} = n - 2a_n. Wait, but earlier we had a_{n+1} = -2a_n + (n - 1). There's a shift in the index. Wait, let's check the index. The recurrence from the work is that for the equation 2a_n + a_{n+1} = n (from equation (i)), so let's set m = n+1, then 2a_{m-1} + a_m = m - 1, so a_m = -2a_{m-1} + m - 1. Which matches the recurrence we had earlier, but with a shift. Yes! If 2a_n + a_{n+1} = n, then for n=2: 2a2 + a3 = 2. n=3: 2a3 + a4 = 3. n=4: 2a4 + a5 = 4. n=5: 2a5 + a6 = 5. n=6: 2a6 + a7 = 6. n=7: 2a7 + a8 = 7. n=8: 2a8 + a9 = 8. n=9: 2a9 + a10 = 9. n=10: 2a10 + a11 = 10. n=11: 2a11 + a12 = 11. Yes! This is the key. The equation (i) is 2a_n + a_{n+1} = n for n ≥ 2. Let's verify with n=2: 2a2 + a3 = 2 ⇒ 20 + (-18) = 2. Correct. n=3: 2a3 + a4 = -36 + 39 = 3. Correct. n=4: 2a4 + a5 = 78 - 74 = 4. Correct. n=5: 2a5 + a6 = -148 + a6 = 5 ⇒ a6 = 153. n=6: 2a6 + a7 = 6 ⇒ 306 + a7 = 6 ⇒ a7 = -300. n=7: 2a7 + a8 = 7 ⇒ -600 + a8 = 7 ⇒ a8 = 607. n=8: 2a8 + a9 = 8 ⇒ 1214 + a9 = 8 ⇒ a9 = -1206. n=9: GameManager + a10 = 9 ⇒ -2412 + a10 = 9 ⇒ a10 = 2421. n=10: 2a10 + a11 = 10 ⇒ 4842 + a11 = 10 ⇒ a11 = -4832. n=11: 2a11 + a12 = 11 ⇒ -9664 + a12 = 11 ⇒ a12 = 9675. Now, let's confirm with the sum formula for n=12. S12 = sum_{k=1}^{12} a_k. Using the formula: S12 = (2/3)a12 + (1/6)(12)^2 - (1/6)(12) + 10 = (2/3)a12 + 24 - 2 + 10 = (2/3)a12 + 32. But also, from the recurrence relation 2a_n + a_{n+1} = n, we can find a relationship between sums. However, the work states that: sum_{k=1}^{12} a_k + sum_{k=1}^5 a_{2k+1} = a1 + a2 + sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) Let's compute the sum inside the brackets: sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) = 2(a3 + a5 + a7 + a9 + a11) + (a4 + a6 + a8 + a10 + a12) From equation (i): 2a_n + a_{n+1} = n, so for n = 2k+1: 2a_{2k+1} + a_{2k+2} = 2k+1. Yes! Because if n = 2k+1, then 2a_n + a_{n+1} = n. So for each k, n = 2k+1, so 2a_{2k+1} + a_{2k+2} = 2k+1. Therefore, sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) = sum_{k=1}^5 (2k + 1) = 2 sum_{k=1}^5 k + sum_{k=1}^5 1 = 2*(15) + 5 = 35. Now, the RHS expression in the work is a1 + a2 + this sum. We have a1 = 7, a2 = 10, so a1 + a2 = 17. Thus, boxed (C) = 17 + 35 = 52. Wait, let's confirm: sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) = sum_{k=1}^5 (2k+1) = (2*1+1) + (2*2+1) +... + (2*5+1) = 3 + 5 + 7 + 9 + 11 = 35. Correct. Then a1 + a2 = 7 + 10 = 17. So total is 17 + 35 = 52. Therefore, boxed (C) = 52. Now, let's confirm with the LHS: sum_{k=1}^{1}^{12} a_k + sum_{k=1}^5 a_{2k+1} = (sum_{k=1}^{12} a_k) + (a3 + a5 + a7 + a9 + a11) But sum_{k=1}^{12} a_k = a1 + a2 + a3 +... + a12 So total LHS = a1 + a2 + 2(a3 + a5 + a7 + a9 + a11) + (a4 + a6 + a8 + a10 + a1 ct 12) = (a1 + a2) + 2(a3 + a5 +... + a11) + (a4 +... + a12) But the sum of the brackets is 2(a3 +... + a11) + (a4 +... + a12) = sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) = 35, so LHS = (a1 + a2) + 35 = 17 + 35 = 52. Correct. Now, we have: - f(n) = n/3 (from boxed (A), which is P(n+1) - P(n) = n/3) - p = boxed (B) = 10 - q = boxed (C) = 52 We need to compute (p * q) / f(12) = (10 * 52) / (12/3) = 520 / 4 = 130. Wait, f(12) = 12/3 = 4. So (10 * 52) / 4 = 520 / 4 = 13. Let's verify all steps: - f(n) = n/3 ⇒ f(12) = 12/3 = 4. - p = 10 (boxed B). - q = 52 (boxed C). - (p * q) / f(12) = (10 * 52) / 4 = 520 / 4 = 13. Yes, that's correct. </think> To solve the problem, we analyze the given sequence and use the provided recurrence relation and summation formula to compute the required expression. --- ### **Step 1: Understanding the Sequence Definition** We are given that for all $ n \geq 2 $,

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

Letusdenote: Let us denote:

S_n = \sum_{k=1}^n a_k.

Then for $ n \geq 2 $,

S_n = \frac{2}{3}a_n + P(n), \quad \text{where } P(n) = \frac{1}{6}n^2 - \frac{1}{6}n + 10.

--- ### **Step 2: Deriving the Recurrence Relation** From the definition, for $ n \geq 2 $:

a_{n+1} = S_{n+1} - S_n = \left( \frac{2}{3}a_{n+1} + P(n+1) \right) - \left( \ for n≥2n \geq 2, we get:

an+1=23(an+1−an)+(P(n+1)−P(n)).a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + (P(n+1) - P(n)).

Compute P(n+1)−P(n)P(n+1) - P(n):

P(n+1)−P(n)=16[(n+1)2−n2]−16[(n+1)−n]=16(2n+1)−16=2n6=n3.P(n+1) - P(n) = \frac{1}{6}[(n+1)^2 - n^2] - \frac{1}{6}[(n+1) - n] = \frac{1}{6}(2n+1) - \frac{1}{6} = \frac{2n}{6} = \frac{n}{3}.

So:

an+1=23(an+1−an)+n3.a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \frac{n}{3}.

Multiplying both sides by 3:

3an+0=2an+1−2an+n⇒2an+an+1=n.3a_{n+0} = 2a_{n+1} - 2a_n + n \Rightarrow 2a_n + a_{n+1} = n.

This matches the identity in the problem:

2an+an+1=3×(A)⇒(A)=n32a_n + a_{n+1} = 3 \times \boxed{(\text{A})} \Rightarrow \boxed{(\text{A})} = \frac{n}{3}.

Thus, f(n)=(A)=n3f(n) = \boxed{(\text{A})} = \frac{n}{3}.

Step 3: Compute Boxed (B)

Given:

∑k=12ak=23a2+16(2)2−16(2)+10.\sum_{k=1}^2 a_k = \frac{2}{3}a_2 + \frac{1}{6}(2)^2 - \frac{1}{6}(2) + 10.

Left-hand side (LHS):

a1+a2=7+a2.a_1 + a_2 = 7 + a_2.

Right-hand side (RHS):

23a2+46−26+10=23a2+26+10=23a2+13+10=23a2+313.\frac{2}{3}a_2 + \frac{4}{6} - \frac{2}{6} + 10 = \frac{2}{3}a_2 + \frac{2}{6} + 10 = \frac{2}{3}a_2 + \frac{1}{3} + 10 = \frac{2}{3}a_2 + \frac{31}{3}.

Equating both sides:

7+a2=23a2+313⇒13a2=313−7=103⇒a2=10.7 + a_2 = \frac{2}{3}a_2 + \frac{31}{3} \Rightarrow \frac{1}{3}a_2 = \frac{31}{3} - 7 = \frac{10}{3} \Rightarrow a_2 = 10.

So, (B)=10\boxed{(\text{B})} = 10.

Step 4: Compute Boxed (C)

We are told:

∑k=112ak+∑k=15a2k+1=a1+a2+∑k=15(2a2k+1+a2k+2).\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}).

Note that from earlier, we have:

2an+an+1=n⇒2a2k+1+a2k+2=2k+1.2a_n + a_{n+1} = n \Rightarrow 2a_{2k+1} + a_{2k+2} = 2k+1.

So:

∑k=15(2a2k+1+a2k+2)=∑k=15(2k+1)=3+5+7+9+11=35.\sum_{k=1}^5 (2a_{2k+1} + a_{2k+2}) = \sum_{k=1}^5 (2k+1) = 3 + 5 + 7 + 9 + 11 = 35.

Also, a1+a2=7+10=17a_1 + a_2 = 7 + 10 = 17.

Thus:

Boxed (C)=17+35=52.\text{Boxed (C)} = 17 + 35 = \boxed{52}.

Step 5: Final Calculation

We now compute:

p×qf(12)=10×52f(12).\frac{p \times q}{f(12)} = \frac{10 \times 52}{f(12)}.

Since f(n)=n3f(n) = \frac{n}{3}, then f(12)=123=4f(12) = \frac{12}{3} = 4.

So:

10×524=5204=13.\frac{10 \times 52}{4} = \frac{520}{4} = \boxed{13}.

✅ Final Answer:

13\boxed{13}
Reasoning Process

[THOUGHT] First, I need to find the values of (A), (B), and (C) from the problem statement.

The sequence is defined with a₁ = 7, and for n ≥ 2, the sum from k=1 to n of aₖ equals (2/3)aₙ + (1/6)n² - (1/6)n + 10.

The problem mentions:

For every natural number n ≥ 2, aₙ₊₁ = sum_{k=1}^{n+1} aₖ - sum_{k=1}^n aₖ, which simplifies to (2/3)(aₙ₊₁ - aₙ) + (boxed A), and after rearranging, 2aₙ + aₙ₊₁ = 3 × (boxed A).

So, boxed A is the expression that when multiplied by 3 gives 2aₙ + aₙ₊₁. Let's derive it properly.

Given the recurrence for the sum, let's compute Sₙ = sum_{k=1}^n a_k. Then Sₙ = (2/3)aₙ + P(n), where P(n) = (1/6)n² - (1/6)n + 10.

For n ≥ 2, Sₙ = (2/3)aₙ + P(n). Also, Sₙ = Sₙ₋₁ + aₙ for n ≥ 2 (since Sₙ = Sₙ₋₁ + aₙ).

So Sₙ₋₁ = Sₙ - aₙ = (2/3)aₙ₋₁ + P(n-1) (for n-1 ≥ 2, i.e., n ≥ 3).

But maybe better to compute aₙ₊₁ = Sₙ₊₁ - Sₙ.

Given Sₙ = (2/3)aₙ + P(n), then Sₙ₊₁ = (2/3)aₙ₊₁ + P(n+1).

Thus, aₙ₊₁ = Sₙ₊₁ - Sₙ = (2/3)aₙ₊₁ + P(n+1) - (2/3)aₙ - P(n).

Rearranging:

aₙ₊₁ - (2/)aₙ₊₁ = - (2/3)aₙ + P(n+1) - P(n)

aₙ₊₁(1 - 2/3) = - (2/3)aₙ + [P(n+1) - P(n)]

aₙ₊₁(1/3) = - (2/3)aₙ + (n/3) [since earlier we saw P(n+1) - P(n) = n/3]

Multiply both sides by 3:

aₙ₊₁ = -2aₙ + n

Wait, but the problem statement says "a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{(\text{A})}" — maybe they did the algebra differently. Let's check the problem's step:

"a_{n+1} = sum_{k=1}^{n+1} a_k - sum_{k=1}^n a_k, so a_{n+1} = (2/3)(a_{n+1} - a_n) + \boxed{(\text{A})}"

Wait, maybe the problem is miswritten here. Let me read the problem again.

"For every natural number n ≥ 2,

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

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

and, after rearranging this expression,

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

Wait, that seems off. Because sum_{k=1}^{n+1} a_k - sum_{k=1}^n a_k = a_{n+1}, which is trivial. But the problem says "a_{n+1} = sum... - sum...", which is just defining a_{n+1} as the difference, but that's always true. But the problem then writes "a_{n+1} = (2/3)(a_{n+1} - a_n) + \boxed{A}".

Wait, maybe there's a mistake here. Let's look back at the sum formula:

sum_{k=1}^n a_k = (2/3)a_n + (1/6)n² - (1/6)n + 10 for n ≥ 2.

So for n and n-1:

sum_{k=1}^n a_k = (2/3)a_n + P(n)

sum_{k=1}^{n-1} a_k = (2/3)a_{n-1} + P(n-1) for n-1 ≥ 2 → n ≥ 3.

Subtracting the two equations: a_n = (2/3)(a_n - a_{n-1}) + P(n) - P(n-1)

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

Then, a_n - (2/3)a_n - (2/3)a_{n-1} = (P(n) - P(n-1))

Wait, no: a_n = (2/3)a_n - (2/3)a_{n-1} + (P(n) - P(n-1))

Bring terms with a_n to left:

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

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

Multiply both sides by 3:

a_n = -2a_{n-1} + 3(P(n) - P(n- osted earlier, P(n) - P(n-1) = (1/6)[(n² - (n-1)^2) - (n - (n-

Reasoning Process

[THOUGHT] Let me carefully work through this problem step by step.

Given:

  • a1=7a_1 = 7
  • For 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

From the problem, we have:

  • an+1=23(an+1−an)+Aa_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{A} (where boxed A = f(n))
  • 2an+an+1=3×A2a_n + a_{n+1} = 3 \times \boxed{A}

Let's derive the recurrence relation properly:

Let Sn=∑k=1nakS_n = \sum_{k=1}^n a_k. Then: Sn=23an+16n2−16n+10S_n = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10

For nn and n−1n-1 (where n≥2n \geq 2): Sn=23an+16n2−16n+10S_n = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10 Sn−1=23an−1+16(n−1)2−16(n−1)+10S_{n-1} = \frac{2}{3}a_{n-1} + \frac{1}{6}(n-1)^2 - \frac{1}{6}(n-1) + 10

Subtracting: Sn−Sn−1=an=23(an−an−1)+[16n2−16n−16(n−1)2+16(n−1)]S_n - S_{n-1} = a_n = \frac{2}{3}(a_n - a_{n-1}) + [\frac{1}{6}n^2 - \frac{1}{6}n - \frac{1}{6}(n-1)^2 + \frac{1}{6}(n-1)] an=23(an−an−1)+16[n2−n−(n2−2n+1)+(n−1)]a_n = \frac{2}{3}(a_n - a_{n-1}) + \frac{1}{6}[n^2 - n - (n^2 - 2n + 1) + (n-1)] an=23(an−an−1)+16[n2−n−n2+2n−1+n−1]a_n = \frac{2}{3}(a_n - a_{n-1}) + \frac{1}{6}[n^2 - n - n^2 + 2n - 1 + n - 1] an=23(an−an−1)+16[2n−2]a_n = \frac{2}{3}(a_n - a_{n-1}) + \frac{1}{6}[2n - 2] an=23(an−an−1)+13(n−1)a_n = \frac{2}{3}(a_n - a_{n-1}) + \frac{1}{3}(n - 1)

But this is for ana_n. We need an+1a_{n+1}, so let's do it for n+1n+1:

Sn+1=23an+1+16(n+1)2−16(n+1)+10S_{n+1} = \frac{2}{3}a_{n+1} + \frac{1}{6}(n+1)^2 - \frac{1}{6}(n+1) + 10 Sn=23an+16n2−16n+10S_n = \frac{2}{3}a_n + \frac{1}{6}n^2 - \frac{1}{6}n + 10

Subtracting: Sn+1−Sn=an+1=23(an+1−an)+16[(n+1)2−(n+1)−n2+n]S_{n+1} - S_n = a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \frac{1}{6}[(n+1)^2 - (n+1) - n^2 + n] an+1=23(an+1−an)+16[n2+2n+1−n−1−n2+n]a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \frac{1}{6}[n^2 + 2n + 1 - n - 1 - n^2 + n] an+1=23(an+1−an)+16[2n]a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \frac{1}{6}[2n] an+1=23(an+1−an)+13na_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \frac{1}{3}n

Rearranging: an+1−23an+1=−23an+13na_{n+1} - \frac{2}{3}a_{n+1} = -\frac{2}{3}a_n + \frac{1}{3}n 13an+1=−23an+13n\frac{1}{3}a_{n+1} = -\frac{2}{3}a_n + \frac{1}{3}n an+1=−2an+na_{n+1} = -2a_n + n

Wait, but this doesn't match the form given in the problem.

Let's use the problem's given form: an+1=23(an+1−an)+Aa_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \boxed{A}

Rearranging: an+1−23an+1=−23an+Aa_{n+1} - \frac{2}{3}a_{n+1} = -\frac{2}{3}a_n + \boxed{A} 13an+1=−23an+A\frac{1}{3}a_{n+1} = -\frac{2}{3}a_n + \boxed{A} an+1=−2an+3×Aa_{n+1} = -2a_n + 3 \times \boxed{A}

Comparing with the derived an+1=−2an+na_{n+1} = -2a_n + n, we get: 3×A=n3 \times \boxed{A} = n A=n3\boxed{A} = \frac{n}{3}

So f(n)=n3f(n) = \frac{n}{3}.

Now for box (B), which corresponds to a2a_2:

For n=2n = 2: ∑k=12ak=23a2+1622−162+10\sum_{k=1}^{2} a_k = \frac{2}{3}a_2 + \frac{1}{6}2^2 - \frac{1}{6}2 + 10 a1+a2=23a2+46−26+10a_1 + a_2 = \frac{2}{3}a_2 + \frac{4}{6} - \frac{2}{6} + 10 7+a2=23a2+26+107 + a_2 = \frac{2}{3}a_2 + \frac{2}{6} + 10 7+a2=23a2+13+107 + a_2 = \frac{2}{3}a_2 + \frac{1}{3} + 10 7+a2=23a2+3137 + a_2 = \frac{2}{3}a_2 + \frac{31}{3} a2−23a2=313−7a_2 - \frac{2}{3}a_2 = \frac{31}{3} - 7 13a2=313−213\frac{1}{3}a_2 = \frac{31}{3} - \frac{21}{3} 13a2=103\frac{1}{3}a_2 = \frac{10}{3} a2=10a_2 = 10

So box (B) = 10.

For box (C), which corresponds to ∑k=112ak+∑k=15a2k+1\sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1}:

From the problem: ∑k=112ak+∑k=15a2k+1=a1+a2+∑k=15(2a2k+1+a2k+2)\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})

From the recurrence 2an+an+1=3×A=n2a_n + a_{n+1} = 3 \times \boxed{A} = n: 2an+an+1=n2a_n + a_{n+1} = n

So for n=2k+1n = 2k+1: 2a2k+1+a2k+2=2k+12a_{2k+1} + a_{2k+2} = 2k+1

Thus: ∑k=15(2a2k+1+a2k+2)=∑k=15(2k+1)=2∑k=15k+∑k=151=2⋅5⋅62+5=30+5=35\sum_{k=1}^{5} (2a_{2k+1} + a_{2k+2}) = \sum_{k=1}^{5} (2k+1) = 2\sum_{k=1}^{5} k + \sum_{k=1}^{5} 1 = 2 \cdot \frac{5 \cdot 6}{2} + 5 = 30 + 5 = 35

So: ∑k=112ak+∑k=15a2k+1=a1+a2+35=7+10+35=52\sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} = a_1 + a_2 + 35 = 7 + 10 + 35 = 52

Thus, box (C) = 52.

Finally, we need to compute p×qf(12)\frac{p \times q}{f(12)} where:

  • p = boxed B = 10
  • q = boxed C = 52
  • f(n) = boxed A = n/3, so f(12) = 12/3 = 4

Thus: p×qf(12)=10×524=5204=13\frac{p \times q}{f(12)} = \frac{10 \times 52}{4} = \frac{520}{4} = 13

Let me verify this calculation using Python. [/THOUGHT]

[PYTHON]

Let's verify the calculation of (p*q)/f(12)

p = 10 # boxed B = a_2 q = 52 # boxed C f_12 = 12 / 3 # f(n) = n/3

result = (p * q) / f_12 result [/PYTHON]

Tool Output
[PYTHON OUTPUT] 130.0 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] Let me carefully recalculate everything step by step:

  1. From the recurrence 2an+an+1=3×(boxed A)2a_n + a_{n+1} = 3 \times \text{(boxed A)}, we have (boxed A)=2an+an+13\text{(boxed A)} = \frac{2a_n + a_{n+1}}{3}.

  2. The problem states that an+1=23(an+1−an)+(boxed A)a_{n+1} = \frac{2}{3}(a_{n+1} - a_n) + \text{(boxed A)}.

  3. From the given sum formula and rearranging, we found that 2an+an+1=n2a_n + a_{n+1} = n, so (boxed A)=n3\text{(boxed A)} = \frac{n}{3}.

  4. Therefore, f(n)=n3f(n) = \frac{n}{3}, so f(12)=123=4f(12) = \frac{12}{3} = 4.

  5. For box (B), we calculated a2=10a_2 = 10 by substituting n=2n = 2 into the sum formula.

  6. For box (C), we calculated ∑k=112ak+∑k=15a2k+1=52\sum_{k=1}^{12} a_k + \sum_{k=1}^{5} a_{2k+1} = 52 using the recurrence relation.

  7. Now we need to compute p×qf(12)=_subsequence10×524=5204=130\frac{p \times q}{f(12)} = \_subsequence{10 \times 52}{4} = \frac{520}{4} = 130.

Wait, I need to be careful. Let me recalculate: 10×524=5204=130\frac{10 \times 52}{4} = \frac{520}{4} = 130

So the answer should be 130.

But let me double-check my calculation: 10 × 52 = 520 520 ÷ 4 = 130

No, that's not right. 4 × 130 = 520, so 520 ÷ 4 = 130

Wait, I'm getting confused. Let me do the calculation properly: 520 ÷ 4 = 130

No, 4 × 130 = 520, so 520 ÷ 4 = 130

Let me use Python to verify:

result = (p * q) / f_12
result

This gives 130.0, which matches the expected calculation.

So the answer is 130\boxed{130}.

Tool Output
[PYTHON OUTPUT] SyntaxError: invalid syntax (<string>, line 1) [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] [PYTHON]

Calculate (p * q) / f(12) where p = 10, q = 52, f(12) = 4

p = 10 q = 52 f_12 = 4

result = (p * q) / f_12 result [/PYTHON]

Tool Output
[PYTHON OUTPUT] 130.0 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python output shows the result is 130.0, which means the final answer is 130.

[FINAL ANSWER] \boxed{130}

Final Model Output
130

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
axk1
10.9

Avg Token Usage

K-EXAONE-236B-A23B
172400.7
Llama-VARCO-8B-Instruct
16930.0
Solar-Open-100B
15251.0
Kanana-2-30B-Thinking-2601
14849.0
Solar-Pro-2 (31B)(high)
13356.0
Kanana-2-30B-Thinking
13274.0
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
K-EXAONE-236B-A23B
6562.0
A.X-4.0 (72B)
6211.0
GPT-5.1 (high)
5774.0
GPT-5.2 (high)
5721.0
axk1
4112.0
EXAONE-4.0.1-32B (high)
940.0