$\sum_{k=0}^n \binom{n}{k}x^k(1-x)^{n-k}k$?

Question

I cant wrap my head around simplifying the following sum:

$$\sum_{k=0}^n \binom{n}{k}x^k(1-x)^{n-k}k,$$ where $0<x<1$.

I tried to apply standard formulas here.

Answer 1

$$(ax+1-x)^n=\sum_{k=0}^n \binom{n}{k}(ax)^k(1-x)^{n-k}$$ then with differentiating respect to $a$ we have $$nx(ax+1-x)^{n-1}=\sum_{k=0}^n \binom{n}{k}a^{k-1}x^k(1-x)^{n-k}k$$ now let $a=1$.

Answer 2

If you see that what you have is the expected value of the Binomial Distribution with n trials probability of success $x$ for any given trial, then we can answer using probabilistic rules.

The binomial distribution is the sum of $n$ independent Bernoulli Distributions of probability of success $x$ in this case. The expected value of any one Bernoulli distribution is $x$. So we have that your sum is equivalent to

$$\underbrace{x+x+x+\cdots+x}_{n\text{ times}} = nx$$

Via the law of the expected value of the sum of random variables.