# Summation of Binomial Expansion with multiplicative factors

Consider the following summation:

$$\sum_{k=0}^n k\binom{n}{k}(1-p)^{n-k}p^k$$

If the $$k$$ term was not present it would be a simple binomial. However, because of the $$k$$ term I am unable to derive the sum.

I believe the summation should be equal to $$np$$ (The summation is part of a larger expression, which upon solving using a different approach gives the $$np$$ result). I have tested this for smaller values of $$n = \{1,2,3\}$$.

Recall that for $$k\geq 1$$, $$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$$ and therefore the sum can be evaluated as a simple binomial: \begin{align} \sum_{k=0}^n k\binom{n}{k}(1-p)^{n-k}p^k &=np\sum_{k=1}^n \binom{n-1}{k-1}(1-p)^{n-1-(k-1)}p^{k-1}\\ &=np((1-p)+p)^{n-1}=np. \end{align} Along the same lines we show the more general identity: $$\sum_{k=0}^n k(k-1)\cdots (k-j)\binom{n}{k}(1-p)^{n-k}p^k =n(n-1)\cdots (n-j)p^{j+1}.$$
Your expression is the mean of a binomial random variable with parameters $$n$$ and $$p$$. This is well known to be $$np$$.