Let $B_n(f)$ be the nth Bernstein Polynomial for $f\in \mathcal{C}[0,1]$ and let $F(x) = x$. Show that $B_n(F)(x)= \sum_{i=0}^{n}F(\frac{i}{n}) {n \choose i} x^i (1-x)^{n-i} = F(x)$.
I began like this $B_n(F)(x)= \sum_{i=0}^{n}F(\frac{i}{n}) {n \choose i} x^i (1-x)^{n-i} = \sum_{i=0}^{n}\frac{i}{n}{n \choose i} x^i (1-x)^{n-i}= \sum_{i=0}^{n}\frac{(n-1)!}{(i-1)!(n-i)!} x^i (1-x)^{n-i}=\sum_{i=0}^{n}{n-1 \choose i-1} x^i (1-x)^{n-i}$.
But now I don't know how to finish this. Any help is appreciated.
We have $B_n(F)(x)=\sum_{k=1}^n\frac{k}{n}\ \binom{n}{k}x^k(1-x)^{n-k}$ and the following implications are verified $$\frac{k}{n}\ \binom{n}{k}=\frac{k}{n}\frac{n!}{k!(n-k)!}=\frac{(n-1)!}{(k-1)![(n-1)-(k-1)]!}=\binom{n-1}{k-1}$$ $$\Rightarrow B_n(F)(x)=x\sum_{k=1}^n\ \binom{n-1}{k-1}x^{k-1}(1-x)^{(n-1)-(k-1)}$$ $$=x\sum_{k=0}^{n-1}\ \binom{n-1}{k}x^{k}(1-x)^{(n-1)-k}=x(x+(1-x))^{n-1}=x.$$ So, $B_n(F)(x)=x.$