Can't find an identity for proving that $ \sum_{k=0}^{i+1} \binom {i+1} k=2^{i+1}$ 
$$ \sum_{k=0}^{i+1} \binom {i+1} k$$

I can't find an identity for this summation :(
To clarify I'm trying to prove using induction that this sum  is equal to $2^{i+1}$, I have my basis and inductive hypothesis done, this is just my inductive step
 A: Your problem can be proved without making direct application of the Binomial Theorem itself. Let us consider two disjoint sets $A$ and $B$ and $n$ elements at our disposal. Now the problem is to divide the $n$ elements in the two sets. In how many ways can it be done? The answer is easy. 
First analyze the problem as for any one of the $n$ elements there are two possibilities of satisfying the condition, namely , either it goes into $A$ or into $B$. It is easy to see that the total number of ways in this case is indeed $2^n$. Right?
Now analyze the same problem a bit differently. Since the sets are disjoint, if we can analyze all the possible distributions in $A$, it will automatically determine all the possible distribution in $B$ because there is a one-one correspondence between the distributions in $A$ and that in $B$. We can take $0$ elements from $n$ elements, the number of ways of choosing in such a manner being $\binom {n}{0}$. Similarly when we choose $1$ element the number of ways becomes $\binom {n}{1}$ and thus generalizing if we choose $i (\leq n)$ elements the number of ways becomes $\binom {n}{i}$ and since each such choice  is mutually exclusive with any other choice the total number of such choices will be $\sum_{i=0}^{n} \binom {n}{i}$. And since the total number of ways must be unique we get $\sum_{i=0}^{n} \binom {n}{i}=2^n$. Hence proved.
