TITLE: Variance of binomial distribution
QUESTION [8 upvotes]: Why for $X\sim B(n,p)$ is $Var(X)=np(1-p)$?
$Var(X)=\sum x_i^2 p_i -(\sum x_i p_i)^2=\sum_{r=0}^n r^2 \binom{n}{r}p^r(1-p)^{n-r}+( \sum_{r=0}^n r \binom{n}{r}p^r(1-p)^{n-r} )^2$
In my short-sightedness, I don't see any viable ways to derive the variance from this.

REPLY [5 votes]: Compute the expected value of $k$
$$
\begin{align}
\mathrm{E}(k)
&=\sum_{k=1}^nk\binom{n}{k}p^k(1-p)^{n-k}\\
&=\sum_{k=1}^nnp\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}\\
&=np(p+(1-p))^{n-1}\\
&=np\tag{1}
\end{align}
$$
Compute the expected value of $k(k-1)$
$$
\begin{align}
\mathrm{E}(k(k-1))
&=\sum_{k=1}^nk(k-1)\binom{n}{k}p^k(1-p)^{n-k}\\
&=\sum_{k=1}^nn(n-1)p^2\binom{n-2}{k-2}p^{k-2}(1-p)^{n-k}\\
&=n(n-1)p^2(p+(1-p))^{n-2}\\
&=n(n-1)p^2\tag{2}
\end{align}
$$
Add $(1)$ and $(2)$ to get $\mathrm{E}(k^2)$ then subtract the square of $(1)$ to get
$$
\begin{align}
\mathrm{Var}(k)
&=\mathrm{E}(k^2)-\mathrm{E}(k)^2\\
&=n^2p^2-np^2+np-n^2p^2\\
&=np(1-p)\tag{3}
\end{align}
$$