Prove the identity $\sum\limits_{k=0}^n\left(x-\frac{k}{n}\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=\frac{1}{n}x(1-x)$ for $0 \leq x \leq 1$ I'd like to prove this identity:
$$\sum_{k=0}^n\left(x-\frac{k}{n}\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=\frac{1}{n}x(1-x)$$
for $x\in[0,1]$ and $n\in\mathbb{N}$. 
I've worked on this problem for the last few hours without success. I have tried induction and I have tried re-writing the expression as many ways as I could think of. Any help would be greatly appreciated. Thanks. 
 A: Brute force approach. There might be a more elegant approach.
Multiply by $n^2$, and you want:
$$\sum_{k=0}^n\left(nx-k\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=nx(1-x)$$
First part:
$$\sum_{k=0}^n n^2x^2\binom{n}{k}x^k(1-x)^{n-k} = n^2x^2(x+(1-x))^n = n^2x^2.\tag{1}$$
Second part:
$$\begin{align}
\sum_{k=0}^n 2nkx \binom{n}{k}x^ky^{n-k} &= 2nx^2\sum_{k=0}^n k\binom{n}{k}x^{k-1}y^{n-k}\\
&=2nx^2\frac{d}{dx}(x+y)^n \\
&= 2nx^2n(x+y)^{n-1}\end{align}$$
Letting $y=1-x$, then we get:
$$\sum_{k=0}^n 2nkx \binom{n}{k}x^k(1-x)^{n-k} = 2n^2x^2\tag{2}$$
Part 3:
$$\begin{align}
\sum k^2\binom{n}{k}x^ky^{n-k} &= x\frac{d}{dx}\sum_{k=0}^nk\binom{n}{k}x^ky^{n-k}\\
&=\left(x\frac{d}{dx}\right)^2(x+y)^n\\
&=x\frac{d}{dx}\left(xn(x+y)^{n-1}\right)\\
&=x\left(n(x+y)^{n-1} + xn(n-1)(x+y)^{n-2}\right)\\
\end{align}$$
Letting $y=1-x$ we have:
$$\sum k^2\binom{n}{k}x^ky^{n-k}=xn + n(n-1)x^2\tag{3}$$
Computing $(1)-(2)+(3)$ we get:
$$\sum_{k=0}^n\left(nx-k\right)^2 \binom{n}{k}x^k(1-x)^k=n^2x^2-2n^2x^2+n(n-1)x^2 + nx=nx-nx^2=nx(1-x)$$
A more algebraic way might amount to noting that:
$$\sum_{k=0}^n ((n-k)x -ky)^2\binom{n}{k}x^ky^{n-k}$$
is a symmetric polynomial in $x,y$, and thus can be written in terms of $x+y,xy$. Not sure where to go from there, however.
