Find explicit formula for summation for p>0 Find explicit formula for summation for any $p>0$:
$$\sum_{k=1}^n(-1)^kk\binom{n}{k}p^k.$$
Any ideas how to do that? I can't figure out how to bite it. I appreciate any help.
 A: Welcome to MSE!
Normally I would flag this as a duplicate, since I'm sure it's been asked before, but approach0 is down right now and some googling around hasn't actually brought it up...
We know from the binomial theorem that
$$\sum_{k=0}^n \binom{n}{k} x^k = (1+x)^n.$$
Differentiating both sides, we find
$$\sum_{k=1}^n k \binom{n}{k} x^{k-1} = n(1+x)^{n-1}$$
If we multiply both sides by $x$ and evaluate at $x = -p$, we find
$$\sum_{k=1}^n k \binom{n}{k} (-p)^k = (-p)n(1-p)^{n-1}$$

I hope this helps ^_^
A: 
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^n}&\color{blue}{(-1)^kk\binom{n}{k}p^k}\\
&=n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}p^k\tag{1}\\
&=n\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}p^{k+1}\tag{2}\\
&\,\,\color{blue}{=-np(1-p)^{n-1}}\tag{3}
\end{align*}

Comment:

*

*In (1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.


*In (2) we shift the index to start with $k=0$.


*In (3) we factor out $-p$ and apply the binomial theorem.
A: \begin{align}
(1-x)^n &= \sum_{k=0}^n \binom{n}{k} (-x)^k
\\
-n(1-x)^{n-1} &= -\sum_{k=1}^n \binom{n}{k} k (-x)^{k-1}
& \text{differentiate w.r.t. $x$}
\\
-nx(1-x)^{n-1} &= \sum_{k=1}^n \binom{n}{k} k (-x)^k
& \text{multiply by $x$}
\end{align}
