How to find $ \binom {1}{k} + \binom {2}{k} + \binom{3}{k} + ... + \binom{n}{k} $ 
Find $$ \binom {1}{k} + \binom{2}{k} + \binom{3}{k} + ... + \binom {n}{k} $$ if $0 \le k \le n$

Any method for solving this problem? I've not achieved anything so far. Thanks in advance!

For $k=0$, the answer is $n$.  
For $k>0$, the answer is ${{n+1}\choose{k+1}}$.
This can be seen by considering how many ways there are to choose $k+1$ integers from the set $\{1,2,\dots,n+1\}$ and conditioning on the greatest integer chosen, because there have to be $k$ choices less than the greatest integer chosen.
For instance we look at ${{n+1}\choose{k+1}}={6\choose 4}$.  To choose $4$ elements from $\{1,2,3,4,5,6\}$, the greatest chosen element could be $4$, and then we would have to choose $3$ elements from $\{1,2,3\}$; or the greatest chosen element could be $5$, and then we would have to choose $3$ elements from $\{1,2,3,4\}$; or the greatest chosen element could be $6$, and then we would have to choose $3$ elements from $\{1,2,3,4,5\}$. 
So altogether, ${6\choose 4}={3\choose 3}+{4\choose 3}+{5\choose 3}$.
Note: the other terms in the sum, of the form ${i\choose k}$ with $i<k$, are all $0$.
