Summation including combination

Question

I am having trouble evaluating the summation:

$\sum_{k=0}^{2n}(-1)^kk^n{2n \choose k}$

Can anyone lead me to a solution?

Also, is there a general or easy way to approach summations that include combinations in them?

Answer 1

We may recognize the action of the forward difference operator. Given a polynomial $p(x)$, $(\delta p)(x)$ is defined as $p(x+1)-p(x)$. We have $(\delta^2 p)(x)=p(x+2)-2p(x+1)+p(x)$ and in general $$ (\delta^m p)(x) = \sum_{k=0}^{m}\binom{m}{k} p(x+k)(-1)^{m-k}. $$ If $p(x)$ is such that $\deg p\geq 1$, we have $\deg(\delta p)=\deg p-1$. Additionally $\delta^{\deg p}p(x) = (\deg p)!$. Now we may consider $p(x)=x^n$, $m=2n$, $x=0$ and draw our conclusion (assuming $n>0$): $$ \sum_{k=0}^{2n} \binom{2n}{k}k^{n}(-1)^k = 0. $$