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.

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 $$r=\left|1+\zeta_4\right|^2=2+\sqrt2.$$
