Question

How do you find the sum of finite geometric series?

Answer 1

`a+ar+ar^2+\cdots+ar^n=\frac{a(r^{n+1}-1)}{r-1}` if `r!=1` and `a+a+a+\cdots+a=(n+1)a` when there are `n+1` terms (and `r=1`).

Explanation:

Consider a finite geometric series with `n+1` terms `a+ar+ar^2+\cdots+ar^n`. Call this sum `S_{n}`. Note that `rS_{n}=ar+ar^2+ar^3+\cdots+ar^{n+1}` so that `rS_{n}-S_{n}=ar^{n+1}-a`.

Solving this equation for `S_{n}` gives `S_{n}=\frac{a(r^{n+1}-1)}{r-1}` when `r!=1`. The number `a` is often referred to as the "first term" and the number `r` is often referred to as the "common ratio". The number `n` is the highest power of `r` in the sum and the sum itself has `n+1` terms.

If `r=1`, then `S_{n}=a+a+a+\cdots+a=(n+1)a`.