# Summand in Cauchy product

The following question arose from general observations for particular cases of $a$ and $b$ when working with infinite MA models.

Let $0 < a, b < 1$. Then the Cauchy product formula yields $$\left( \sum_{k=0}^\infty a^k \right) \left( \sum_{j=0}^\infty b^j \right)= \sum_{k=0}^\infty \sum_{i=0}^k a^i b^{k-i}$$

Mathematica suggests $$\sum_{i=0}^k a^i b^{k-i} = \frac{a^{k+1} - b^{k+1}}{a - b},$$ but I have no idea how to prove the result.

You can prove $\sum_{i=0}^k a^i b^{k-i} = \frac{a^{k+1} - b^{k+1}}{a - b}$ by induction !