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I'm trying to understand the following the proof.

I want to show that

$$E\left[\frac{1}{X+1}\right] = \frac{1}{(n+1)p}(1-(1-p)^{n+1})$$

The proof goes like this:

$$ \begin{align} E\left[\frac{1}{X+1}\right] & = \sum\limits_{i=0}^n \frac{n!}{(n-i)!(i+1)!} p^i (1-p)^{n-i} \\[6pt] & = \frac{1}{(n+1)p} \sum\limits_{i=0}^n \frac{(n+1)!}{(n-i)!(i+1)!} p^{i+1} (1-p)^{n-i} \\[6pt] & = \frac{1}{(n+1)p} \sum\limits_{i=0}^n \frac{(n+1)!}{(n-i)!(i+1)!} p^{i+1} (1-p)^{n-i} \\[6pt] & = \frac{1}{(n+1)p}\sum\limits_{i=1}^{n+1} \frac{(n+1)!}{(n+1-i)!(i)!} p^{i} (1-p)^{n+1-i} \\[6pt] & = \frac{1}{(n+1)p}(1-(1-p)^{n+1}) \end{align} $$

I understand the proof until the very last step. But then I can't see how (or why) all the i terms cancel out. Can someone help me understand this?

Thanks

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1 Answer 1

up vote 1 down vote accepted

That's because $\sum\limits_{i=1}^{n+1}a_i=(\sum\limits_{i=0}^{n+1}a_i)-a_0$ where $a_i=\binom{n+1}{i}p^i(1-p)^{n+1-i}.$ The binomial sum is $\sum\limits_{i=0}^{n+1} \binom{n+1}{i}p^i(1-p)^{n+1-i}=(p+(1-p))^{n+1}=1$ and $a_0=(1-p)^{n+1}.$

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