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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?

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

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  • $\begingroup$ That's a $2n$-fold iterated difference. As long as $n>0$, this one is zero. $\endgroup$ – Lord Shark the Unknown May 17 at 16:06
  • $\begingroup$ 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. $\endgroup$ – Adrian Keister May 17 at 16:08
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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 $n>0$): $$ \sum_{k=0}^{2n} \binom{2n}{k}k^{n}(-1)^k = 0. $$

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