I am struggling to simplify this series: $$\sum_{k = 0}^{\infty} \left(\!\!{n\choose k}\!\!\right) a^k k^2 = \sum_{k = 0}^{\infty} {n + k - 1\choose k}\ a^k k^2.$$
I understand that the square under the sum makes it harder, but I was wondering if there exists a simplification nevertheless.
Please, also write if you know any tricks that may help to reduce the power of $k$ under the summation, as without $k^2$ the series evaluates to $(1 - a)^{-n}$.
Because $$k\left(\!\!\binom{n}{k}\!\!\right) = n\left(\!\!\binom{n+1}{k-1}\!\!\right),$$ we have \begin{align} \sum_{k=0}^\infty \left(\!\!\binom{n}{k}\!\!\right) a^k k^2 &= \sum_{k=1}^\infty \left(\!\!\binom{n}{k}\!\!\right) a^k k^2 \\ &= n \sum_{k=1}^\infty \left(\!\!\binom{n+1}{k-1}\!\!\right) a^k k \\ &= n \sum_{k=1}^\infty \left(\!\!\binom{n+1}{k-1}\!\!\right) a^k (k-1+1) \\ &= n \sum_{k=2}^\infty \left(\!\!\binom{n+1}{k-1}\!\!\right) a^k (k-1) + n \sum_{k=1}^\infty \left(\!\!\binom{n+1}{k-1}\!\!\right) a^k \\ &= n(n+1) \sum_{k=2}^\infty \left(\!\!\binom{n+2}{k-2}\!\!\right) a^k + n \sum_{k=1}^\infty \left(\!\!\binom{n+1}{k-1}\!\!\right) a^k \\ &= a^2n(n+1) \sum_{k=0}^\infty \left(\!\!\binom{n+2}{k}\!\!\right) a^k + an \sum_{k=0}^\infty \left(\!\!\binom{n+1}{k}\!\!\right) a^k \\ &= a^2n(n+1) \frac{1}{(1-a)^{n+2}} + an \frac{1}{(1-a)^{n+1}} \\ &= \frac{an(an+1)}{(1-a)^{n+2}}. \end{align}