Prove that $\sum^{n}_{k=0}(x-\frac{k}{n})^{2}{{n}\choose{k}}x^{k}(1-x)^{n-k}=\frac{x(1-x)}{n}$ holds for ${0}\leq{x}\leq{1}$

Prove that
$\sum^{n}_{k=0}(x-\frac{k}{n})^{2}{{n}\choose{k}}x^{k}(1-x)^{n-k}=\frac{x(1-x)}{n}$
holds for ${0}\leq{x}\leq{1}$

For ${0}\leq{x}\leq{1}$, prove that the following identity
holds:$$\sum^{n}_{k=0}(x-\frac{k}{n})^{2}{{n}\choose{k}}x^{k}(1-x)^{n-k}=\frac{x(1-x)}{n}$$