Show that $\sum_{r=0}^{n} \binom{n}{r} r p^r q^{n-r} = np$, given $p+q=1$

Question

How can I prove

$$\sum_{r=0}^{n} \binom{n}{r} {r} p^r q^{n-r} = np,$$ given that $p+q=1$?

I think it is about applying $$(1+x)^n=\sum_{r=0}^{n} \binom{n}{r} x^r.$$ Although I am quite unsure about the approach, I think it might give a way.

Answer 1

For $n\ge r\ge1,$ $$r\binom nr=n\binom{n-1}{r-1}$$

$$\implies\sum_{r=0}^{n} \binom{n}{r} {r} p^r q^{n-r} =np\sum_{r=1}^n\binom{n-1}{r-1}p^{r-1}q^{n-1-(r-1)}=np(p+q)^{n-1}$$

Answer 2

Another approach uses calculus. Since $\sum_{r\ge 0}\binom{n}{r}x^r=(1+x)^n$, $$\sum_{r\ge 0}\binom{n}{r}rx^r=x\frac{d}{dx}(1+x)^n=nx(1+x)^{n-1}.$$Set $x=\frac{p}{q}$ then multiply by $q^n$, giving $$\sum_r\binom{n}{r}rp^rq^{n-r}=npq^{n-1}\left(1+\tfrac{p}{q}\right)^{n-1}=np(p+q)^{n-1}=np.$$

Answer 3

Here is a probabilistic approach:

• $\sum_{r=0}^{n} \binom{n}{r} r p^r q^{n-r} = E[X]$, where
• $X$ follows a binomial distribution $B(p,n)$, which means
• $X = \sum_{i=1}^nX_i$ with $X_i$ i.i.d. where $X_i \in \{0,1\}$ and $P(X_i=1) = p$
• $\Rightarrow E[X_i] = 1\cdot p + 0 \cdot (1-p) = p$

Now, it follows $$E[X] = E \left[\sum_{i=1}^nX_i\right] = \sum_{i=1}^n E[X_i] =\sum_{i=1}^n p = np$$