Prove Value of Summation with Combinatoric

Question

Prove that:

$$\sum_{k=0}^n\binom{n}{k}\left(\frac{k}{n}\right)^2x^k(1-x)^{n-k}=\frac{x(1+x(n-1))}{n}$$

I've managed to simplify the sum up until this point:

$$\frac{\sum_{k=1}^n\binom{n-1}{k-1}kx^k(1-x)^{n-k}}{n}$$

But I'm not sure where to go from here. It looks like I need to get it into a form where I can apply Binomial Theorem, but I'm not entirely sure how to do that since there's that $k$ term in there.

Answer 1

HINT

Use that

$$\left(\frac{k}{n}\right)^2\binom{n}{k}=\left(\frac{k\left(k-1\right)}{n^2}+\frac{k}{n^2}\right)\binom{n}{k}=\frac{n-1}{n}\frac{k\left(k-1\right)}{n\left(n-1\right)}\binom{n}{k}+\frac{1}{n}\frac{k}{n}\binom{n}{k}$$

Hence

$$ \left(\frac{k}{n}\right)^2\binom{n}{k}=\frac{n-1}{n}\binom{n-2}{k-2}+\frac{1}{n}\binom{n-1}{k-1}$$ And here appears the binomial theorem you mentioned.