# Calculating a recursive power term binomial sum

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

I have been trying for a few hours now and I start thinking it may be not possible to find the answer directly, I also think it is equal to: $$x^n+(-2)^{n}n!.$$ The right answer is in fact: $$(-2)^{n}n!.$$

Thank you very much,

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As you think you have found a closed form, you could try to show it by induction. –  azimut Mar 31 '13 at 16:00
Are you sure it's equal? A quick calculation with anything like $n=2$ and $x=2$ for example produces different values. –  Guest 86 Mar 31 '13 at 16:05

Hint: The summand $(x-2(k+1))^n$ is $(-2k)^n$ plus a polynomial in $k$ of degree less than $n$. Now use the formulas here to show that your expression is $(-2)^n n!$, (without $x$).

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Thank you very much, I will try like that when I have a moment! –  Nre Mar 31 '13 at 18:51

Suppose we seek to evaluate $$\sum_{k=0}^n {n\choose k} (-1)^{n-k} (x-2(k+1))^n.$$

Introduce $$(x-2(k+1))^n = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((x-2(k+1))z) \; dz.$$

This yields for the sum $$\frac{n!\times (-1)^n}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \sum_{k=0}^n {n\choose k} (-1)^k \exp((x-2(k+1))z) \; dz \\ = \frac{n!\times (-1)^n}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((x-2)z) \sum_{k=0}^n {n\choose k} (-1)^k \exp(-2kz) \; dz \\ = \frac{n!\times (-1)^n}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((x-2)z) (1-\exp(-2z))^n \; dz.$$

Now the series for $1-\exp(-2z)$ starts at $2z$ and hence the residue times the factor is $$n!\times (-1)^n\times 2^n.$$

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