proof of formula and calculation sum

Question

Show that following formula is true:

$$ \sum_{i=0}^{[n/2]}(-1)^i (n-2i)^n{n \choose i}=2^{n-1}n! $$

Using formula calculate $$ \sum_{i=0}^n(2i-n)^p{p \choose i} $$

Answer 1

We show by inclusion/exclusion that $$\sum_{i=0}^n (-1)^i (c-k i)^n \binom ni = k^n n!$$ for any integer $n$ and complex $c$. (Honestly!) Given this, take $c=n$ and $k=2$, note that in this case the term with $i=n/2$ vanishes (if there is one), and your formula follows.

Here's the inclusion/exclusion argument. First suppose $c>k n$ is an integer, and let $S_1,\ldots, S_n$ be disjoint $k$-tuples in $[c]=\{1,\ldots,c\}$. We count all functions from $[n]$ to $[c]$ such that the range of $f$ intersects each $S_j$. On the one hand, there are clearly $k^n n!$ such functions (for each $j$, there's a unique $i$ such that $f(i)\in S_j$, and there are $k$ possibilities for $f(i)$ within $S_j$). On the other hand, the number of functions $f:[n]\to [c]$ which miss a given collection of $i$ of the $S_j$'s is $(c-k i)^n$, so we're done by inclusion/exclusion.

It's clear that the answer doesn't depend on $c$, as long as $c$ is sufficiently large. But the left-hand side is a finite polynomial in $c$, so it must be a constant polynomial, and therefore you can pick $c$ arbitrarily.

Answer 2

Suppose we seek to evaluate

$$S_1 = \sum_{q=0}^{\lfloor n/2 \rfloor} {n\choose q} (-1)^q (n-2q)^n.$$

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

and furthermore introduce $$[[0\le q \le \lfloor n/2 \rfloor]] = \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1+w+w^2+\cdots+w^{\lfloor n/2 \rfloor}}{w^{q+1}} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{w^{\lfloor n/2 \rfloor+1}-1}{(w-1)w^{q+1}} \; dw.$$

This is an Iverson bracket that ensures that we may extend the range of the sum from $\lfloor n/2 \rfloor$ to $n.$

We get for the sum $$\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{w^{\lfloor n/2 \rfloor+1}-1}{(w-1)w} \\ \times \sum_{q=0}^{n} {n\choose q} (-1)^q \exp((n-2q)z) \frac{1}{w^q} \; dw \; dz \\ = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{w^{\lfloor n/2 \rfloor+1}-1}{(w-1)w} \\ \times \sum_{q=0}^{n} {n\choose q} (-1)^q \exp(-2qz) \frac{1}{w^q} \; dw \; dz.$$

This is $$\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{w^{\lfloor n/2 \rfloor+1}-1}{(w-1)w} \left(1-\frac{\exp(-2z)}{w}\right)^n \; dw \; dz \\ = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{w^{\lfloor n/2 \rfloor+1}-1}{(w-1)w^{n+1}} \left(w-\exp(-2z)\right)^n \; dw \; dz .$$

Call this integral $J_1.$

An alternate representation of the sum is $$S_2 = (-1)^n \sum_{q=\lfloor n/2 \rfloor+1}^n {n\choose q} (-1)^q (2q-n)^n.$$

The Iverson bracket now becomes $$[[\lfloor n/2 \rfloor+1\le q\le n]] = \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{w^{\lfloor n/2 \rfloor+1}+\cdots+w^n}{w^{q+1}} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\epsilon} w^{\lfloor n/2 \rfloor+1} \frac{w^{n-\lfloor n/2 \rfloor}-1}{(w-1)w^{q+1}} \; dw.$$

We get for the sum $$(-1)^n \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} w^{\lfloor n/2 \rfloor+1} \frac{w^{n-\lfloor n/2 \rfloor}-1}{(w-1)w} \\ \times \sum_{q=0}^{n} {n\choose q} (-1)^q \exp((2q-n)z) \frac{1}{w^q} \; dw \; dz \\ = (-1)^n \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(-nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} w^{\lfloor n/2 \rfloor+1} \frac{w^{n-\lfloor n/2 \rfloor}-1}{(w-1)w} \\ \times \sum_{q=0}^{n} {n\choose q} (-1)^q \exp(2qz) \frac{1}{w^q} \; dw \; dz.$$

This is $$(-1)^n \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(-nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} w^{\lfloor n/2 \rfloor+1} \frac{w^{n-\lfloor n/2 \rfloor}-1}{(w-1)w} \\ \times \left(1-\frac{\exp(2z)}{w}\right)^n \; dw \; dz \\ = (-1)^n \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(-nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} w^{\lfloor n/2 \rfloor+1} \frac{w^{n-\lfloor n/2 \rfloor}-1}{(w-1)w^{n+1}} \\ \times \left(w-\exp(2z)\right)^n \; dw \; dz .$$

Call this integral $J_2.$

By cancellation of the $w^{n+1}$ factor this is $$- (-1)^n \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(-nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(w-1)w^{n-\lfloor n/2 \rfloor}} \left(w-\exp(2z)\right)^n \; dw \; dz .$$

Now put $z=-v$ to get $$(-1)^n \frac{n!}{2\pi i} \int_{|v|=\epsilon} \frac{\exp(nv)}{(-1)^{n+1} v^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(w-1)w^{n-\lfloor n/2 \rfloor}} \\ \times\left(w-\exp(-2v)\right)^n \; dw \; dv \\ = - \frac{n!}{2\pi i} \int_{|v|=\epsilon} \frac{\exp(nv)}{v^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(w-1)w^{n-\lfloor n/2 \rfloor}} \\ \times\left(w-\exp(-2v)\right)^n \; dw \; dv .$$

It follows that $$J_1+J_2 = 2S \\ = -\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(w-1)w^{n+1}} \left(w-\exp(-2z)\right)^n \; dw \; dz .$$

This is $$-\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(w-1)w^{n+1}} \\ \times \sum_{q=0}^n {n\choose q} (w-1)^{n-q} (1-\exp(-2z))^q \; dw \; dz.$$

Now we have two cases namely $q=n$ and $q\lt n$. When $q=n$ we get the integral $$-\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(w-1)w^{n+1}} (1-\exp(-2z))^n \; dw \; dz.$$

The series for $1-\exp(-2z)$ starts with $2z$ so that on extracting the residue in $z$ we obtain $$-\frac{n!\times 2^n}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(w-1)w^{n+1}} \; dw \\ = \frac{n!\times 2^n}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(1-w)w^{n+1}} \; dw = n! \times 2^n.$$

It remains to show that the following integral is zero: $$-\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{(w-1)w^{n+1}} \\ \times \sum_{q=0}^{n-1} {n\choose q} (w-1)^{n-q} (1-\exp(-2z))^q \; dw \; dz.$$

This is $$-\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp(nz)}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \\ \times \sum_{q=0}^{n-1} {n\choose q} (w-1)^{n-q-1} (1-\exp(-2z))^q \; dw \; dz$$

which is seen to be zero by inspection since the maximum power of the $w-1$ term is $n-1$ and we are extracting the $w^n$ coefficient.

We have shown that $2S = 2^n \times n!$ and hence $$S = 2^{n-1} \times n!$$ as claimed.

A similar technique was used at this MSE link I and this MSE link II.

Answer 3

By way of commentary on the answer by @Tad it appears in retrospect that the key is to realize that

$$\sum_{q=0}^{\lfloor n/2 \rfloor} {n\choose q} (-1)^q (n-2q)^n = \sum_{q=\lfloor n/2 \rfloor+1}^n {n\choose q} (-1)^q (n-2q)^n$$

which is trivial once you see it (reindex $q$ by $n-q$) but escaped my attention as I worked on the problem.

Of course the sum

$$\sum_{q=0}^n (-1)^q (c-kq)^n {n\choose q}$$ is easy to evaluate algebraically (complex variables).

Just introduce $$(c-kq)^n = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((c-kq)z) \; dz.$$

This yields for the sum $$\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \sum_{q=0}^n {n\choose q} (-1)^q \exp((c-kq)z) \; dz \\ = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp(cz) \sum_{q=0}^n {n\choose q} (-1)^q \exp(-kqz) \; dz \\ = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp(cz) \left(1-\exp(-kz)\right)^n \; dz.$$

Since $1-\exp(-kz)$ starts at $kz$ the residue is $k^n$ for a final answer of $k^n \times n!.$

The defect of my first answer was that I did not see the symmetry in the sum.