<s>[INST] 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!
 [/INST] No formal answer, but a nice illustration:
The solution can exemplarically be shown by a matrix-multiplication-scheme. The following shows $S \cdot P = X$, where $X$ is simply a shifting of $P$ .
The left-bottom is matrix $S$ which performs the partial summations of the columns of matrix $P$ which is in the right-top-matrix, and the result $X$ is the right-bottom-matrix:
$$ \small \begin{matrix} 
 . & . & . & . & . & | & 1 & . & . & . & . & | \\ 
 . & . & . & . & . & | & 1 & 1 & . & . & . & | \\ 
 . & . & . & . & . & | & 1 & 2 & 1 & . & . & | \\ 
 . & . & . & . & . & | & 1 & 3 & 3 & 1 & . & | \\ 
 . & . & . & . & . & | & 1 & 4 & 6 & 4 & 1 & | \\ 
 - & - & - & - & - & + & - & - & - & - & - & + \\ 
 1 & . & . & . & . & | & 1 & . & . & . & . & | \\ 
 1 & 1 & . & . & . & | & 2 & 1 & . & . & . & | \\ 
 1 & 1 & 1 & . & . & | & 3 & 3 & 1 & . & . & | \\ 
 1 & 1 & 1 & 1 & . & | & 4 & 6 & 4 & 1 & . & | \\ 
 1 & 1 & 1 & 1 & 1 & | & 5 & 10 & 10 & 5 & 1 & | \\ 
 - & - & - & - & - & + & - & - & - & - & - & +
 \end{matrix} $$
From this a hypothetical result can easily be formulated and gives the direction of the attempt of a proof... 
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