@Secret We first need to get information about the polynomials $r_p(x)={}_n C_px^p(1-x)^{n-p}$ that appear in the expansion. Note that ${}_nC_p$ are just the binomial coefficients. Skipping some fiddling with the binomial formula, we have $$\sum_{p=0}^n r_p(x)=1,\quad \sum_{p=0}^n pr_p(x)=nx,\quad \sum_{p=0}^n p(p-1)r_p(x)=n(n-1)x^2.$$
Using this, we find $$\sum_{p=0}^n (p-nx)^2r_p(x)=nx(1-x).$$
Since $f\in C[0,1]$ is continuous and $[0,1]$ is compact, we suppose $|f(x)|\le M$ in $[0,1]$. We may also assume $f$ is uniformly continuous, i.e. for $\epsilon>0$ there is a $\delta>0$ such that …