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 a polynom $P(x)=a_mx^m +\dots +a_1x+a_0$ denote $[x^k]P(x)=a_k$.
Notice that for $k\le i \le n$
$$\dbinom{i}{k}=[x^k](1+x)^i$$
hence
\begin{align}
\sum_{i=1}^n\dbinom{i}{k}&=\sum_{i=k}^n\dbinom{i}{k}\\
&=\sum_{i=1}^n[x^k](1+x)^i\\
&=[x^k]\sum_{i=k}^n(1+x)^i\\
&=[x^k]\frac{(1+x)^{n+1}-(1+x)^k}{(1+x)-1}\\
&=[x^k]\frac{(1+x)^{n+1}-(1+x)^k}{x}\\
&=[x^{k+1}]\left((1+x)^{n+1}-(1+x)^k\right)\\
&=[x^{k+1}](1+x)^{n+1}+[x^{k+1}](1+x)^{k}\\
&=\dbinom{n+1}{k+1}+0\\
&=\dbinom{n+1}{k+1}
\end{align}
