I saw a proof for Vandermonde's identity: Proof for Vandermonde's identity: ${{m+n} \choose r} = \sum_{k=0}^r {m \choose k}{n\choose {r-k}}$
I'm new to working with sigmas, so I tried to understand the sigma transition they did, but when I thought I got it, I tried to implement it on a different double sigma but it did not work:
\begin{equation}\begin{aligned} \sum_{i=0}^3i\sum_{j=0}^3j=\sum_{i=0}^3\sum_{j=0}^3(ij)=36 \end{aligned}\end{equation} Then i did what they did by setting k=i+j, and by adding 3+3 the the outer sigma: \begin{equation}\begin{aligned} \sum_{k=0}^6\sum_{i=0}^ki(k-i)=70 \end{aligned}\end{equation} Is there something i'm missing here? how should you do this transition to relate i and j? Because i've seen these formulas in algorithm analysis (with variables instead of numbers so I can't calculate them) but I don't really know how to solve these problems.
