Find explicit formula for summation for p>0

Question

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.

Answer 1

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

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

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I hope this helps ^_^

Answer 2

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.

Answer 3

\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}