3
$\begingroup$

This question already has an answer here:

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

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4
$\begingroup$

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 $$ n(x+(1-x))^{n-1} = nx. $$

$\endgroup$
  • $\begingroup$ Hello. I edited the question. It should be $$x^t(1-x)^{n-t}$$. I apologize. I was sleepy when I asked this question. $\endgroup$ – Caesar Ian Pedro Apr 21 '17 at 10:52
  • $\begingroup$ Answer updated. $\endgroup$ – Chappers Apr 21 '17 at 12:13
3
$\begingroup$

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 $x$.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.