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?
 A: 
We obtain
\begin{align*}
\sum_{k=0}^n\binom{n+1}{k+1}&=\sum_{k=1}^{n+1}\binom{n+1}{k}\tag{1}\\
&=\sum_{k=0}^{n+1}\binom{n+1}{k}-1\tag{2}\\
&=(1+1)^{n+1}-1\tag{3}\\
&=2^{n+1}-1
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we shift the index of the sum by one to start from $k=1$.

*In (2) we add the summand with index $k=0$ and subtract $1$ correspondingly.

*In (3) we apply the binomial theorem.
A: Both sides count the number of subset of a set with $n+1$ elements, minus the empty set. Also, the formula could be rewritten as:
$$\sum_{k=1}^n\left(\matrix{n\\k}\right) = 2^n -1$$
A: Use inductive approach
For n = 0: $$\binom{1}{1} = 1 = 2^1 - 1$$ => TRUE
Assume that for n>=0, the following (*) is true $$\sum_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$$ 
We need prove: $$\sum_{k=0}^{n+1} \binom{n+2}{k+1} = 2^{n+2} - 1$$
Remember: $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$
So: $$\binom{n+2}{k+1} = \binom{n+1}{k+1} + \binom{n+1}{k}$$
=> $$\sum_{k=0}^{n+1} \binom{n+2}{k+1} = \sum_{k=0}^{n+1}\binom{n+1}{k+1}+\sum_{k=0}^{n+1}\binom{n+1}{k}$$
$$=\binom{n+1}{n+2} + \sum_{k=0}^{n}\binom{n+1}{k+1} + \sum_{k=0}^{n}\binom{n+1}{k+1} - \binom{n+1}{0}$$
$$=2*\sum_{k=0}^{n}\binom{n+1}{k+1} - 1 = 2*2^{n+1} - 1 = 2^{n+2}-1$$
That is what to be proven.
A: 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  $
