what is the value of K for given recurrence relation?

Question

Consider the recurrence relation

$a_1 = 8$, and for all $n>1$, $a_n=6n^2 + 2n+a_{n–1}$. Let $a_{99} = K\times 10000$. The value of $K$ is ___.

Answer 1

Note that, $a_n=a_1+\sum\limits_{i=2}^n{6i^2+2i}=8+n(n+1)(2n+1)-6+n(n+1)-2=2n(n+1)^2$

So, $a_{99}=2 \times 99 \times 100^2=198 \times 10000$

Hence, your $K$ is equal to $198$.

Answer 2

Hint: $a_{99}=a_1+\displaystyle\sum_{n=2}^{99}(a_n-a_{n-1})$