$\sum_{k=0}^{n-1}a^k=(a^n-1)/(a-1)$

Let $a$ be a real number other than 1. Use induction to show that for each positive integer $n$, \[\sum_{k=0}^{n-1} a^k = (a^n-1)/(a-1).\]

Solution

For $n=1$, \[ \sum_{k=0}^{0}a^k=1=\frac{a-1}{a-1}. \] Assume \[ \sum_{k=0}^{n-1}a^k=\frac{a^n-1}{a-1}. \] Then \[ \sum_{k=0}^{n}a^k =\frac{a^n-1}{a-1}+a^n =\frac{a^{n+1}-1}{a-1}. \] Thus the formula holds for all positive integers $n$ by induction.