<s>[INST] How to prove that $\sum\limits_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$ using the Binomial Theorem? I have this proposition:
$$\sum_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$$
How can I prove that? How to use the Binomial Theorem to solve that?
 [/INST] Show that $\sum\limits_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$
Solution:
$\sum\limits_{k=0}^{n} \binom{n}{k} = \sum\limits_{k=0}^{n}\binom{n}{k+1} + \binom{n}{0}$
$\sum\limits_{k=0}^{n} \binom{n}{k} = \sum\limits_{k=0}^{n}\binom{n}{k+1} +1$
therefrore,
$\sum\limits_{k=0}^{n} \binom{n}{k+1} =\sum\limits_{k=0}^{n} \binom{n}{k} -1  $ 
$\sum\limits_{k=0}^{n+1} \binom{n+1}{k+1} =\sum\limits_{k=0}^{n+1} \binom{n+1}{k} -1  $ .....(1)
we know,
$\sum\limits_{k=0}^{n} \binom{n}{k} = 2^{n} $
so,
$\sum\limits_{k=0}^{n+1} \binom{n+1}{k} = 2^{n+1} $
finally (1) can be written as...
$\sum\limits_{k=0}^{n+1} \binom{n+1}{k+1} =2^{n+1} -1  $
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