I want to compute $S(n,m,a)=\sum_{k=0}^{n}k^{m}\cdot\binom{n}{k}\cdot a^k.$ With $n,m\in\mathbb N$, $a\neq0$ and $S(n,0,a)=(a+1)^n$.
What I have found already:
I don't see any other options then integrating until we've got Newton's formula and then differentiating it as many times as we integrated.
I have found some values:$S(n,1,a)=na\cdot(a+1)^{n-1}$ $S(n,2,a)=an(an+1)(a+1)^{n-2}$ $S(n,3,a)=an(a^2n^2+3an-a+1)(a+1)^{n-3}.$
And, since $\displaystyle\sum_{k=0}^{n}k^{m}\cdot\binom{n}{k}a^k=a\cdot\frac{\mathrm d}{\mathrm da}\int\displaystyle\sum_{k=0}^{n}k^{m}\cdot\binom{n}{k}a^{k-1}\mathrm da=a\cdot\frac{\mathrm d}{\mathrm da}\sum_{k=0}^{n}k^{m-1}\cdot\binom{n}{k}a^k$, I have the recursion $S(n,m,a)=a\cdot\frac{\mathrm d}{\mathrm da}S(n,m-1,a).$ But I can't find any general formula.