GATE 2016 CS – Question 37
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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Correct answer: 197.9 to 198.1
Explanation
Unrolling the recurrence, $a_n = 8 + \sum_{k=2}^{n}(6k^2 + 2k)$. The $k = 1$ term would be $6 + 2 = 8$, so this equals $\sum_{k=1}^{n}(6k^2 + 2k)$. For $n = 99$: $6 \times \frac{99 \cdot 100 \cdot 199}{6} + 2 \times \frac{99 \cdot 100}{2} = 1970100 + 9900 = 1980000$. So $K \times 10^4 = 1980000$ and $K = 198$.