Evaluate $\sum_{r=0}^{100} {(-1)^{r} {100 \choose r} {r^{50}}}$

Question

The problem is to the evaluate the following sum:

$$\sum_{r=0}^{100} {(-1)^{r} {100 \choose r} {r^{50}}}$$

I don't see how I can get started on this, since the function attached to the combination is polynomial, not an exponential.

I also tried to look for a combinatorial approach, using the inclusion exclusion principle. I guess it can be interpreted as the arrangement of 100 distinct items in 50 places, however the answer doesn't seem to be correct.

Would love some help on this one!

Answer 1

HINT:

You can show that for any $0\le k \le 50$ we have $$\sum_{r\ge 0} (-1)^r \binom{100}{r}\cdot \binom{r}{k} = 0$$

Then check that you can write $r^{50}$ uniformly in $r$ as a linear combination of $\binom{r}{k}$ for $0\le k \le 50$.

Answer 2

Evaluating

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

we find

$$\sum_{r=0}^{2n} (-1)^r {2n\choose r} n! [z^n] \exp(rz) \\ = n! [z^n] \sum_{r=0}^{2n} (-1)^r {2n\choose r} \exp(rz) \\ = n! [z^n] (1-\exp(z))^{2n}$$

Note however that $1-\exp(z) = - z + \cdots$ and hence $(1-\exp(z))^{2n}$ starts with $z^{2n}$, showing that the sum is zero.