Sum $(1-x)^n$ $\sum_{r=1}^n$ $r$ $n\choose r$ $(\frac{x}{1-x})^r$

Question

The question is to find the value of:

$n\choose 1$$x(1-x)^{n-1}$ +2.$n\choose2$$x^2(1-x)^{n-2}$ + 3$n\choose3$$x^3(1-x)^{n-3}$ .......n$n\choose n$$x^n$.

I wrote the general term and tried to sum it as:

S=$(1-x)^n$$\sum_{r=1}^n$$r$$n\choose r$$(\frac{x}{1-x})^r$.

I got stuck here.

What do I do after this?

Answer 1

HINT: $r\binom{n}r=n\binom{n-1}{r-1}$; this is easy to see if you expand into factorials, and it also has a straightforward combinatorial proof.

Answer 2

Note that your sum can also be written as $$ \begin{align} \sum_{r=1}^nr\binom{n}{r}x^r(1-x)^{n-r} &=nx\sum_{r=1}^n\binom{n-1}{r-1}x^{r-1}(1-x)^{n-r}\\ &=nx(x+1-x)^{n-1}\\[9pt] &=nx \end{align} $$ This is the expected value of a binomial distribution.