TITLE: Help understanding the weak law of large numbers with respect to statistics
QUESTION [11 upvotes]: I'm trying to do some self-studying to upgrade my statistics knowledge, and came across this term in a section discussing the weak law of large numbers and Bernoulli's theorem:
$$\sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$$
According to the book I was reading, this term can "easily be shown to equal np". 
I am at a loss as to how to do so and could use some guidance!
Edited
Lower limit $k=0$, not $1$

REPLY [2 votes]: Let $X_n=B(n,p)$ be a binomially distributed random variable. Also notice that $X_n=Y_1+Y_2+\cdots+ Y_n$ where $Y_i$ are i.i.d. Bernoulli with parameter $p$.
Now observe that
\begin{align}
\sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}&= \operatorname{E}(X_n)\\
&= \operatorname{E}(  Y_1+Y_2+\cdots Y_n)\\
&=\operatorname{E}(  Y_1)+\operatorname{E}(Y_2)+\cdots +\operatorname{E}(Y_n)\\
&=np
\end{align}