prove $$\sum_{j=i}^n \begin{pmatrix} j-1 \\ i-1 \end{pmatrix} = \begin{pmatrix} n \\ i \end{pmatrix}$$ What I am doing now is to time $i$ on both side and try to argue out of this: $$\sum_{j=i}^n i \begin{pmatrix} j-1 \\ i-1 \end{pmatrix} = i\begin{pmatrix} n \\ i \end{pmatrix}$$


marked as duplicate by JMoravitz, Clayton, Trevor Gunn, peterwhy, Guido A. Oct 7 '18 at 1:14

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