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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.

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  • $\begingroup$ combine the alternating part and p exponentiated part and you will get something that looks like the binomial theorem. Maybe you can combine the coefficient of k into the binomial coefficient, aka the combination function? $\endgroup$ – Tyma Gaidash May 6 at 19:19
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    $\begingroup$ Hint. Note that $k\binom{n}{k}=n\binom{n-1}{k-1}$ for $1\leq k\leq n$ $\endgroup$ – Robert Z May 6 at 19:20
  • $\begingroup$ Can you give me another hint? I still don't see what to do next after this step. $\endgroup$ – TheStudent21 May 6 at 19:24
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    $\begingroup$ Look at $(1+x)^n=\sum_{k=0}^{n}\binom{n}{k}x^k$ and differentiate both sides. $\endgroup$ – Benjamin St. May 6 at 19:24
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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 ^_^

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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.

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\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}

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