Expectation of the binomial distribution

Question

In the proof of the expectation of the binomial distribution,

$$E[X]=\sum_{k=0}^{n}k \binom{n}{k}p^kq^{n-k}=p\frac{d}{dp}(p+q)^n=pn(p+q)^{n-1}=np$$

Why is $\sum_{k=0}^{n}k \binom{n}{k}p^kq^{n-k}= p\frac{d}{dp}(p+q)^n$?

I know that by the binomial theorem $(p+q)^n=\sum_{k=0}^{n}\binom{n}{k}p^kq^{n-k}$.

Answer 1

You're missing the sum over $k$ in the binomial theorem. If you add it and perform the differentiation and multiplication termwise, you'll get the desired result.

Answer 2

$$\begin{align}\mathsf E(X)&=\sum_{k=0}^n k\tbinom nkp^kq^{n-k} \\[1ex]&=\sum_{k=0}^n\tbinom nk p(kp^{k-1})q^{n-k} \\[1ex]&=\sum_{k=0}^n\tbinom nkp(\tfrac{\partial~~}{\partial p}p^k)q^{n-k} \\[1ex]&=p\tfrac{\partial~~}{\partial p}\sum_{k=0}^n\tbinom nkp^kq^{n-k} \\[1ex]&=p\tfrac {\partial~~}{\partial p}(p+q)^n \\[1ex]&=np(p+q)^{n-1} \\[1ex]&=np\end{align}$$