Prove that (a) for each $n \ge 1$ $$\sum_{k=0}^n C_{n}^{k} \left(\frac{k}{n}-\frac{1}{2} \right)^2 \frac{1}{2^n}=\frac{1}{4n}$$(b) for every n \ge m \ge 2 $$\sum_{\ell=0}^n \sum_{k_1+...+k_n=\ell,k_i=0,...,m} \frac{\ell!}{k_1!...k_n!} \frac{1}{(m+1)^n} \left(\frac{\ell}{n}-\frac{m}{2} \right)^2= \left(\frac{m^3-3m^2}{12(m+1)}+\frac{m}{2}-\frac{m}{3(m+1)}\right)n$$