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?

Let $b_{n}=\frac{a_{n}}{2}$. 
Then $b_{n}=3n^2+n+b_{n-1}$. 
For any $n$th degree polynomial $Q(x)$, one can find a $n+1$th degree polynomial $P(x)$ such that $P(x)-P(x-1)=Q(x)$. 
Then, use this to note that $b_{n}-n(n+1)^2=b_{n-1}-(n-1)n^2$, since $n(n+1)^2-(n-1)n^2=3n^2+n$. 
Thus $b_{n}-n(n+1)^2$ is a constant. But since $b_{1}=0$, then this implies that $a_{n}=2n(n+1)^2$. 
Thus $a_{99}=198 \times 10^4$. 
