Here are some simple proofs based on the binomial theorem. They also demonstrate how amazing mathematical proofs can be.

## 1. The hockey stick identity

By the formula for the sum of a geometric series, \(\frac{(1+r)^{n+1}-1}{(1+r)-1}=\displaystyle\sum_{i=0}^n (1+r)^i\). Now expanding both sides by the binomial theorem and simplifying, $$\displaystyle\sum_{i=0}^n \binom{n+1}{i+1}r^i=\displaystyle\sum_{i=0}^n \displaystyle\sum_{j=0}^i \binom{i}{j}r^j$$ Now rearrange the sum on the left (it is a finite sum after all) to get $$\displaystyle\sum_{i=0}^n \binom{n+1}{i+1}r^i=\displaystyle\sum_{j=0}^n r^j\displaystyle\sum_{i=j}^n \binom{i}{j}$$ Equating coefficients of \(r^k\) on both side gives the dsired result.

## 2. Vandermonde’s identity

Well, obviously, \((1+r)^n (1+r)^m =(1+r)^{n+m}\). Again, use the binomial theorem to expand both sides: $$\left(\displaystyle\sum_{i=0}^n \binom{n}{i}r^i \right)\left(\displaystyle\sum_{i=0}^m \binom{m}{i}r^i \right)=\displaystyle\sum_{i=0}^{n+m} \binom{n+m}{i}r^i$$ The coefficient of \(r^k\) in the LHS is $$\displaystyle\sum_{i+j=k}\binom{n}{i}\binom{m}{j} =\displaystyle\sum_{i=0}^k \binom{n}{i}\binom{m}{k-i}$$ so equating coeffiecients of \(r^k\) yields Vandermonde’s identity.

## 3. Tri-Vandermonde

Same as #2, except with the product of three binomials.