<s>[INST] Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let  $a_{99}=K\times10^4$ .The value of $K$ is______ .

My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+6.1^2+2.1$
$=6(n^2+(n−1)^2+...+2^2+1^2)+2(n+(n−1)+...+2+1)$
$=6×n(n+1)(2n+1)/6+2×n(n+1)/2$
$=n(n+1)(2n+1+1)$
$=2n^3+2n^2+2n^2+2n$
$=2n(n^2+n+n+1)$
$=2n(n^2+2n+1)$
$a_n=2n(n+1)^2$
for $n=99, a_{99}=2×99×(99+1)^2=198×10^4$

I'm looking for short trick or alternative way, can you explain please?
 [/INST] $$a_n-a_{n-1}=6n^2+2n$$
Now sum both sides from $n=2$ to $x$. You end up with a telescoping sum on the left hand side which will drastically make things easier and help you find the closed form, assuming you know some summation formulas.
$$\sum_{n=2}^x (a_n-a_{n-1})= \sum_{n=2}^x (6n^2+2n)$$
$$a_x-a_1=\sum_{n=2}^x (6n^2+2n)$$
$$a_x=a_1+\sum_{n=2}^x (6n^2+2n)$$
$$a_x=a_1-8+\sum_{n=1}^x (6n^2+2n)$$
$$a_x=\sum_{n=1}^x (6n^2+2n)$$
Therefore $a_{99}$ would be:
$$a_{99}=\sum_{n=1}^{99} (6n^2+2n)$$
Where now you can apply summation formulas.
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