<s>[INST] Why Does $ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} $ sum to $ (1-(1-p)^{n+1}) $? 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!
 [/INST] 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)
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