Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1. Let a99 = K × 104. The value of K is …

Official answer: 198 (±0.1). Explanation: Given the recurrence relation for n ≥ 2: an - an-1 = 6n2 + 2n Notice that for n = 1, 6(1)2 + 2(1) = 8 = a1. Therefore, we can express an as: an = Σk=1n (6k2 + 2k) = 6 Σk=1n k2 + 2 Σk=1n k Using the standard summation formulas: Σk=1n k2 = (n(n+1)(2n+1))/(6) Σk=1n k = (n(n+1))/(2) Substitute these into the expression for an: an = 6 · (n(n+1)(2n+1))/(6) + 2 · (n(n+1))/(2) an = n(n+1)(2n+1) + n(n+1) = n(n+1)(2n + 1 + 1) = 2n(n+1)2 For n = 99: a99 = 2 × 99 × (99 + 1)2 = 198 × 1002 = 198 × 104 Since a99 = K × 104, we have: K = 198

Combinatorics

GATE CSE 2016 Set 1 · Q27

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Consider the recurrence relation $a_1 = 8$ , $a_n = 6n^2 + 2n + a_{n-1}$ . Let $a_{99} = K \times 10^4$ . The value of $K$ is ________ .

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Official answer: 198 (±0.1) (±0.1)

Combinatorics

Given the recurrence relation for $n \ge 2$ :

a_n - a_{n-1} = 6n^2 + 2n

Notice that for $n = 1$ , $6(1)^2 + 2(1) = 8 = a_1$ . Therefore, we can express $a_n$ as:

a_n = \sum_{k=1}^{n} (6k^2 + 2k) = 6 \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k

Using the standard summation formulas:

\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

\sum_{k=1}^{n} k = \frac{n(n+1)}{2}

Substitute these into the expression for $a_n$ :

a_n = 6 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}

a_n = n(n+1)(2n+1) + n(n+1) = n(n+1)(2n + 1 + 1) = 2n(n+1)^2

For $n = 99$ :

a_{99} = 2 \times 99 \times (99 + 1)^2 = 198 \times 100^2 = 198 \times 10^4

Since $a_{99} = K \times 10^4$ , we have:

K = 198

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This Combinatorics previous year question is Q27 of the GATE CSE 2016 Set 1 paper. Find more GATE CSE PYQs on Combinatorics in Engineering Mathematics, or browse all GATE CSE PYQs.

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