Calculate sum $\sum\limits_{k=0}k^2{{n}\choose{k}}3^{2k}$. I need to find calculate the sum Calculate sum $\sum\limits_{k=0}k^2{{n}\choose{k}}3^{2k}$.
Simple algebra lead to this $\sum\limits_{k=0}k^2{{n}\choose{k}}3^{2k}=n\sum\limits_{k=0}k{{n-1}\choose{k-1}}3^{2k}$. But that's still not very helpful. This binomial screws everything up for me, I would like a nice recurrence relation, but don't know what to do with it.

We have $\displaystyle k^2=k(k-1)+k$
So, $\displaystyle k^2 \binom nk=k(k-1)\binom nk+k\binom nk$
Now $\displaystyle k\cdot\binom nk=k\frac{n!}{(n-k)!k!}=kn\frac{(n-1)!}{[n-1-(k-1)]!(k-1)!\cdot k}=n\binom{n-1}{k-1}$
and  $\displaystyle k(k-1)\cdot\binom nk=k(k-1)\frac{n!}{(n-k)!k!}=k(k-1)n(n-1)\frac{(n-2)!}{[n-2-(k-2)]! (k-2)!\cdot k(k-1)}=n(n-1)\binom{n-2}{k-2}$
$\displaystyle\implies\sum_{k=0}^n k^2 \binom nk3^{2k}=\sum_{k=0}^n k^2 \binom nk9^k$
$\displaystyle=9n \sum_{k=0}^n\binom{n-1}{k-1}9^{k-1}+n(n-1)9^2\sum_{k=0}^n\binom{n-2}{k-2}9^{k-2}$
Utilize $\binom mr=0$ for $r<0$ or $r>m$
More generally, $\displaystyle \sum_{r=0}^ma_rk^r=b_0+b_1\cdot k+b_2\cdot k(k-1)+\cdots+b_{m-1}\prod_{r=0}^{m-2}(k-r)+b_m\prod_{r=0}^{m-1}(k-r)$ where $b_r$s are arbitrary constants 
