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I was browsing around when I found this question: Find the expected value of $\frac{1}{X+1}$ where $X$ is binomial.

I understood the solution until I hit this portion where $ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} $ becomes $ (1-(1-p)^{n+1}) $...

Any help would be appreciated. Thanks!

Community
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crentagon
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2 Answers2

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Change the summation index to get:

$ \sum\limits_{k=1}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} p^{k} (1-p)^{n+1-k} $.

If you add to that the summand corresponding to $k=0$, i.e. $(1-p)^{n+1}$, you have

$ \sum\limits_{k=0}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} p^{k} (1-p)^{n+1-k} = [p+(1-p)]^{n+1} = 1^{n+1}= 1$.

So, your expression is $1$ minus what you added, i.e. $1 - (1-p)^{n+1}$.

Hypergeometricx
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quid
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By using $t = k + 1$:

$$ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} = \sum\limits_{t = 1}^{n+1} \begin{pmatrix} n+1 \\ t \end{pmatrix} p^{t} (1-p)^{n-t +1}$$

$$ = \sum\limits_{t = 0}^{n+1} \begin{pmatrix} n+1 \\ t \end{pmatrix} p^{t} (1-p)^{n-t +1} - (1 - p)^{n+1} $$

$$=\sum\limits_{t = 0}^{n+1} \begin{pmatrix} n+1 \\ t \end{pmatrix} p^{t} (1-p)^{(n+1)-t } -(1 - p)^{n+1} = (p + 1 - p)^{n+1} - (1-p)^{n+1} = 1 - (1 - p)^{n+1}$$

(Using the binomial theorem)