# Evaluating a summation related to binomial expansion [duplicate]

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How do you evaluate this summation $$\sum_{t=0}^{n}{t\binom{n}{t}x^t(1-x)^{n-t}}$$ where $0 < x < 1?$

## marked as duplicate by user21820, YuiTo Cheng, RRL, Xander Henderson, postmortesJun 26 at 16:29

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## 2 Answers

The easiest way is to write $1-x=a$ for the time being, and then your sum is $$\sum_{t=0}^n \binom{n}{t}t x^t a^{n-t} = x\frac{d}{dx}\sum_{t=0}^n \binom{n}{t} x^ta^{n-t} = x\frac{d}{dx} (x+a)^n = nx(x+a)^{n-1},$$ and returning to $1-x$ gives $$n(x+(1-x))^{n-1} = nx.$$

• Hello. I edited the question. It should be $$x^t(1-x)^{n-t}$$. I apologize. I was sleepy when I asked this question. – Caesar Ian Pedro Apr 21 '17 at 10:52
• Answer updated. – Chappers Apr 21 '17 at 12:13

A variation of the theme:

We obtain \begin{align*} \sum_{t=0}^nt\binom{n}{t}x^t(1-x)^{n-t}&=n\sum_{t=1}^n\binom{n-1}{t-1}x^t(1-x)^{n-t}\tag{1}\\ &=nx\sum_{t=0}^{n-1}\binom{n-1}{t}x^t(1-x)^{n-1-t}\tag{2}\\ &=nx \end{align*}

Comment:

• In (1) we use $\binom{n}{t}=\frac{n}{t}\binom{n-1}{t-1}$ and we also start with index $t=1$, since the term with $t=0$ is zero.

• In (2) we shift the index $t$ by one to start from $t=0$ and factor out $x$.

• Hello. I edited the question. It should be $$x^t(1-x)^{n-t}$$. I apologize. I was sleepy when I asked this question. – Caesar Ian Pedro Apr 21 '17 at 10:52
• @CaesarIanPedro: Answer updated accordingly. Regards, – Markus Scheuer Apr 21 '17 at 10:56