GATE CS 2016 Set 1 Q.37 (CS Q.27): Combinatorics, Counting, Recurrence Relations and Generating Functions

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 _____.

Your answer

Official answer and solution

Solution

Express the recurrence as a summation and evaluate using formulas for $\sum n^2$ and $\sum n$ .

The recurrence relation is:

a_n - a_{n-1} = 6n^2 + 2n \quad \text{for } n \ge 2

with base condition $a_1 = 8$ .

Notice that for $n = 1$ , $6(1)^2 + 2(1) = 8 = a_1$ . Therefore, assuming $a_0 = 0$ , we can express $a_n$ as:

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

Substitute the standard summation formulas:

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

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

Evaluate for $n = 99$ :

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

We are given $a_{99} = K \times 10^4$ . Equating gives $K = 198$ .

Final answer

198

Common trap

Summing from $i = 1$ to 99 and adding $a_1 = 8$ a second time.

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