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?

As $a_n-a_{n-1}$ is a quadratic polynomial, $a_n$ must be a cubic polynomial. By integration, close to $\dfrac63n^3=2n^3$.
Then the factor $10^4$ is very likely to be $(99+1)^2$, and we can gamble on the expression $2n(n+1)^2$. It works for $n=1$ ($a_1=8$) and $n=2$ ($a_2=36$). This is convincing enough.
$$2n=198.$$

By better knowledge of Faulhaber's formula, (the first coefficient is always $\dfrac1d$, the second coefficient is always $\dfrac12$), we can predict that the cubic term is exactly $\dfrac63n^3=2n^3$ and the quadratic term is exactly $\dfrac62n^2+n^2=4n^2$. This fits with $2n(n+1)^2$.
