# Summation including combination [closed]

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?

## closed as off-topic by Nosrati, YuiTo Cheng, Cesareo, José Carlos Santos, Xander HendersonMay 18 at 21:08

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• That's a $2n$-fold iterated difference. As long as $n>0$, this one is zero. – Lord Shark the Unknown May 17 at 16:06
• Wolfram Programming Lab (Mathematica) gives $(-1)^{2n} (2n)! \operatorname{StirlingS2}[n, 2n],$ which as Lord Shark pointed out appears to evaluate to $0$ if $n>0.$ The StirlingS2[n,m] function gives the number of ways of partitioning a set of $n$ elements into $m$ non-empty subsets. Clearly, you can't partition a set of $n$ elements into $2n$ non-empty subsets, hence the zero value. – Adrian Keister May 17 at 16:08

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 $$n>0$$): $$\sum_{k=0}^{2n} \binom{2n}{k}k^{n}(-1)^k = 0.$$