Sum of $\sum_{0\leq k\leq n}k\binom{n}{k}p^k(1-p)^{n-k}$

Question

I would like to evaluate the sum $$\sum_{k = 0}^{n}k\binom{n}{k}p^k(1-p)^{n-k} = \sum_{k}k\binom{n}{k}p^k(1-p)^{n-k}$$

(With $n$ integer)

This looks very close to $\sum_{k}\binom{n}{k}p^k(1-p)^{n-k}$, which is of course equal to $1$ (from Binomial Theorem, or probabilistic interpretation and most likely thousands of other reasons). How can the additional factor be handled?

Answer 1

As for $k\ge1,$ $$k\binom nk=\cdots=n\binom{n-1}{k-1}$$

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

Answer 2

$\dfrac {d}{dx}(y+x)^n = n(y+x)^{n-1}=$

$(1/x) \sum_{k=0}^{n} k \binom{n}{k} y^{n-k}x^{k} $.

$y:=1-p; x:= p.$

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