二項係数の級数（かんたん）

$\zeta =\exp \left(\frac{2\pi i}{k}\right),i=\sqrt{-1}$とします.

$$\begin{array}{rl}& \sum _{l=0}^{\lfloor \frac{n-r}{k}\rfloor }(\genfrac{}{}{0ex}{}{n}{kl+r})\\ & =\frac{1}{k}\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n}{l})\left(\sum _{j=0}^{k-1}{\zeta }^{j(l-r)}\right)\\ & =\frac{1}{k}\sum _{l=0}^{k-1}{\zeta }^{-rl}\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n}{j}){\zeta }^{lj}\\ & =\frac{1}{k}\sum _{l=0}^{k-1}{\zeta }^{-rl}(1+{\zeta }^l{)}^n\\ & =\frac{1}{k}\sum _{l=0}^{k-1}{\zeta }^{-rl}\sqrt{2^n{(1+\cos \frac{2l}{k}\pi )}^n}{(\frac{1+\cos \frac{2l}{k}\pi }{\sqrt{2(1+\cos \frac{2l}{k}\pi )}}+i\frac{\sin \frac{2l}{k}\pi }{\sqrt{2(1+\cos \frac{2l}{k}\pi )}})}^n\\ & =\frac{1}{k}\sum _{l=0}^{k-1}{\zeta }^{-rl}{2}^n|{\cos }^n\frac{l}{k}\pi |{(|\cos \frac{l}{k}\pi |+i\mathrm{sgn}(\sin \frac{2l}{k}\pi )|\sin \frac{l}{k}\pi |)}^n\\ & =\frac{2^n}{k}(1+\sum _{0<l<\frac{k}{2}}{\zeta }^{-rl}|{\cos }^n\frac{l}{k}\pi |(\cos \frac{nl}{k}\pi +i\sin \frac{nl}{k}\pi )+\sum _{\frac{k}{2}<l<k}{\zeta }^{-rl}|{\cos }^n\frac{l}{k}\pi |(-1{)}^n(\cos \frac{nl}{k}\pi +i\sin \frac{nl}{k}\pi ))\\ & =\frac{2^n}{k}(1+\sum _{0<l<\frac{k}{2}}|{\cos }^n\frac{l}{k}\pi |(\cos \frac{(n-2r)l}{k}\pi +i\sin \frac{(n-2r)l}{k}\pi )+(-1{)}^n|{\cos }^n\frac{k-l}{k}\pi |(\cos \frac{(n-2r)(k-l)}{k}\pi +i\sin \frac{(n-2r)(k-l)}{k}\pi ))\\ & =\frac{2^n}{k}(1+2\sum _{0<l<\frac{k}{2}}|{\cos }^n\frac{l}{k}\pi |\cos \frac{(n-2r)l}{k}\pi )\\ & =\frac{2^n}{k}(1+2\sum _{0<l<\frac{k}{2}}{\cos }^n\frac{l}{k}\pi \cos \frac{(n-2r)l}{k}\pi )\\ & =\frac{2^n}{k}\sum _{l=0}^{k-1}{\cos }^n\frac{l}{k}\pi \cos \frac{(n-2r)l}{k}\pi \end{array}$$