# Sums of binomial coefficients

Does anyone know something about the following sums?
$$S_m(n)=\sum\limits_{k=o}^n(-1)^k{mn\choose mk}$$
Notice that $S_m(n)=0$ for odd $n$, so we only consider $S_m(2n)$. It holds that $S_0(2n)=1$, $S_1(2n)=0$, $S_2(2n)=(-4)^n$, $S_3(2n)=\frac{2}{3}(-27)^n$, but $S_4(2n)$ is no longer a geometric progression.

#### Solutions Collecting From Web of "Sums of binomial coefficients"

For every sequence $(u(k))_k$, consider $$T_N(u)=\sum\limits_{k=0}^N{N\choose k}u(k),$$ and note that $S_m(2n)=T_{2mn}(u_m)$ where $u_m$ is defined by $u_m(k)=1$ if $2m$ divides $k$, $u_m(k)=-1$ if $m$ divides $k$ but $2m$ does not, and $u_m(k)=0$ for every other $k$.
The sequence $(u_m(k))_k$ has period $2m$ hence there exists some coefficients $(c_j^m)$ such that, for every $k$, $$u_m(k)=\sum_{j=1}^{2m}c_j^m\zeta_m^{kj},$$ where $\zeta_m=\exp(\mathrm i\pi/m)$ is the primitive $2m$-th root of unity. This yields $$T_N(u_m)=\sum_{j=1}^{2m}c_j^m\sum\limits_{k=0}^N{N\choose k}\zeta_m^{kj}=\sum_{j=1}^{2m}c_j^m(1+\zeta_m^j)^N,$$ hence $$S_m(2n)=\sum_{j=1}^{2m}c_j^m(1+\zeta_m^j)^{2mn},\qquad\zeta_m=\exp(\mathrm i\pi/m).$$
One can note that the mean of $u_m$ is zero hence $c_{2m}^m=0$.

For example, if $m=4$, then $$4u(m)=\zeta_4^{k}+\zeta_4^{3k}+\zeta_4^{5k}+\zeta_4^{7k},\qquad\zeta_4=(1+\mathrm i)/\sqrt2,$$
hence all this leads (I think) to $$S_4(2n)=\frac{1+(-1)^nr^{4n}}2,$$ where $$r=\left|1+\zeta_4\right|^2=2+\sqrt2.$$

Start by restating the problem: we seek to evaluate
$$S_m(n) = \sum_{k=0}^n (-1)^k {nm\choose km} = \sum_{k=0}^n (-1)^k {nm\choose nm-km}.$$

Introduce the integral representation
$${nm\choose nm-km} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm-km+1}} \; dz.$$

This gives the following for the sum:
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}} \sum_{k=0}^n \frac{(-1)^k}{z^{-km}} \; dz.$$

We may extend the sum to infinity because the integral correctly
represents the binomial coefficient being zero for $k>n$ to get
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}} \sum_{k\ge 0} \frac{(-1)^k}{z^{-km}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}} \sum_{k\ge 0} (-1)^k z^{km} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{nm}}{z^{nm+1}} \frac{1}{1+z^m}\; dz.$$

Extracting coefficients we obtain
$$[z^{nm}] (1+z)^{nm} \frac{1}{1+z^m} = \sum_{q=0}^{nm} {nm\choose nm-q} [z^q] \frac{1}{1+z^m} = \sum_{q=0}^{nm} {nm\choose q} [z^q] \frac{1}{1+z^m}.$$

Introduce $\rho_k = e^{i\pi/m + 2\pi ik/m}$ so that using partial
fractions by residues on simple poles we have
$$\frac{1}{1+z^m} = \sum_{k=0}^{m-1} \frac{\mathrm{Res}\left(\frac{1}{1+z^m}; z=\rho_k\right)}{z-\rho_k}.$$
This is
$$\sum_{k=0}^{m-1} \frac{\frac{1}{m} \rho_k^{-(m-1)}}{z-\rho_k} = \sum_{k=0}^{m-1} \frac{\frac{1}{m} \rho_k^{-m}}{z/\rho_k-1}.$$

Substituting this into the sum formula yields
$$\sum_{q=0}^{nm} {nm\choose q} [z^q] \sum_{k=0}^{m-1} \frac{\frac{1}{m} \rho_k^{-m}}{z/\rho_k-1} = – \sum_{q=0}^{nm} {nm\choose q} \sum_{k=0}^{m-1} \frac{1}{m} \rho_k^{-m} \rho_k^{-q}$$

Now $\rho_k^{-m} = -1$ so this yields
$$\sum_{q=0}^{nm} {nm\choose q} \sum_{k=0}^{m-1} \frac{1}{m} \rho_k^{-q} = \frac{1}{m} \sum_{k=0}^{m-1} \sum_{q=0}^{nm} {nm\choose q} \rho_k^{-q} \\ = \frac{1}{m} \sum_{k=0}^{m-1} \left(1 + \frac{1}{\rho_k}\right)^{nm} = \frac{1}{m} \sum_{k=0}^{m-1} \left(1 + \rho_k\right)^{nm}.$$

A trace as to when this method appeared on MSE and by whom starts at this