Any other ways to evaluate $\displaystyle \sum_{k=0}^n k {n \choose k} p^k (1-p)^{n-k}$?

Question

$\displaystyle \sum_{k=0}^n k {n \choose k} p^k (1-p)^{n-k}$ with $0<p<1$

I know of one way to evaluate it (from statistics) but I was wondering if there are any other ways.

This is the way I know:

Let

$$M(t)=\displaystyle \sum_{k=0}^n e^{kt} {n \choose k} p^k (1-p)^{n-k}$$

Then $$M(t)=\displaystyle \sum_{k=0}^n {n \choose k} (pe^t)^k (1-p)^{n-k}=(pe^t+1-p)^n$$

$$M'(t)=\displaystyle \sum_{k=0}^n ke^{kt} {n \choose k} p^k (1-p)^{n-k}=pe^tn(pe^t+1-p)^{n-1}$$

$$M'(0)=\displaystyle \sum_{k=0}^n k {n \choose k} p^k (1-p)^{n-k}=np$$

Answer 1

Yes.

$$\begin{align} \sum_{k=0}^n k \binom{n}{k} p^k (1-p)^{n-k} &= \sum_{k=1}^n k \binom{n}{k} p^k (1-p)^{n-k} = \sum_{k=1}^n k\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}\\ &=\sum_{k=1}^n \frac{n!}{(k-1)!(n-1-(k-1))!} p^k (1-p)^{n-k}\\ &=np\sum_{k=1}^n \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} p^{k-1} (1-p)^{n-1-(k-1)}\\ &=np\sum_{\ell=0}^{n-1} \frac{(n-1)!}{\ell!(n-1-\ell)!} p^{\ell} (1-p)^{n-1-\ell}\\ &=np\sum_{\ell=0}^{n-1} \binom{n-1}{\ell} p^{\ell} (1-p)^{n-1-\ell}\\ &= np. \end{align}$$

Answer 2

You can also notice that this is the expectation $E(X)$ where $X$ is a binomial random variable with parameters $n$ and $p$, and $E(X)=n\,p$.

Answer 3

A slightly different variation of an answer already given which might also be convenient.

We obtain \begin{align*} \color{blue}{\sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k}} &=np(1-p)^{n-1}\sum_{k=1}^n\binom{n-1}{k-1}\left(\frac{p}{1-p}\right)^{k-1}\tag{1}\\ &=np(1-p)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}{k}\left(\frac{p}{1-p}\right)^{k}\tag{2}\\ &=np(1-p)^{n-1}\left(1+\frac{p}{1-p}\right)^{n-1}\tag{3}\\ &\color{blue}{=np}\tag{4} \end{align*}

Comment:

• In (1) we use the binomial identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and we factor out $p(1-p)^{n-1}$ to prepare the index shift of the next line.

• In (2) we shift the index to start from $k=0$.

• In (3) we use the binomial summation formula.

• In (4) we do some final simplifications.