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

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.
