I need to calculate the following sum:
$\sum_{x=0}^y x\binom{y}{x}\alpha^x(1-\alpha)^{y-x}$
i know it equals: $\alpha y$, but I don't know how to calculate it exactly.
My steps: $\sum_{x=0}^y x\binom{y}{x}\alpha^x(1-\alpha)^{y-x}=\sum_{x=0}^y \frac{x\alpha^x(1-\alpha)^{y-x}\cdot y!}{x!(y-x)!}=\sum_{x=0}^y \frac{\alpha^x(1-\alpha)^{y-x}\cdot y!}{(x-1)!(y-x)!}$
Is it correct?
$$ \begin{aligned} \sum_{x=0}^yx {y \choose x}\alpha^x(1-\alpha)^{y-x} &= \sum_{x=1}^yx \frac{y!}{x!(y-x)!}\alpha^x(1-\alpha)^{y-x} = \\ &= \sum_{x=1}^y\frac{y!}{(x-1)!(y-x)!}\alpha^x(1-\alpha)^{y-x} = \\ &= \alpha y\sum_{x=1}^y \frac{(y-1)!}{(x-1)!\left((y-1)-(x-1)\right)!}\alpha^{x-1}(1-\alpha)^{(y-1)-(x-1)} = |x{-}1 \rightarrow x| = \\ &= \alpha y \sum_{x=0}^{y-1}\frac{(y-1)!}{x!\left((y-1)-x\right)!}\alpha^x(1-\alpha)^{(y-1)-x} = \\ &= \alpha y\underbrace{\sum_{x=0}^{y-1}{y-1 \choose x}\alpha^x(1-\alpha)^{(y-1)-x}}_{=\left(\alpha + (1-\alpha)\right)^{y-1} = 1} = \alpha y \end{aligned} $$
