Bound on the sum with binomial coefficients

Question

Let $m \in N$. How to bound from above and below the following sum: $$ \sum_{k=0}^n{n \choose k}a^k b^{n-k}k^m $$

(I know that $\sum_{k\geq 0}{n \choose k}a^k b^{n-k}=(a+b)^n$, but what happened when under summation we multiply by extra $k^m$?)

Answer 1

Let $$S_m=\sum_{k=0}^{n} k^m a^k*b^{n-k}$$ By binomial theorem we havr $$(a+b)^n=\sum_{k=0}^{m} a^k b^{n-k}$$ Differentiating w.r.t. $a$ we can get $$S_1=an(a+b)^{n-1}$$ Again d.w.r.t. a, we get $$S_2=an(a+b)^{n-2}(an+b)$$ Next d.w.r.t $a$ gives $$S_3=an(a+b)^{n-3}[a^2 n^2+3abn+b^2-ab]$$ Next, we get $$S_4=an(a+b)^{n-4} P_3(n),$$ where $P_3(n)$ is a polynomial of degree $n$. A compact general result is difficult to come by.