Exercise 251Logic, sets and proofs › Proof techniques › InductionHigh schoolAdded 3 Nov 2026

Geometric sum by induction

#induction#geometric series#sum

Statement

Prove by induction: for $r\neq1$ and $n\ge1$ , $\sum_{k=0}^{n-1}r^k=\frac{1-r^n}{1-r}$ .

Hint

Base case $n=1$ ; inductive step: add $r^n$ .

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Solution

Base case. $n=1$ : $r^0=1=\frac{1-r}{1-r}$ . ✓

Inductive step. Assume the formula. Then
$\sum_{k=0}^{n}r^k=\frac{1-r^n}{1-r}+r^n=\frac{1-r^n+r^n(1-r)}{1-r}=\frac{1-r^n+r^n-r^{n+1}}{1-r}=\frac{1-r^{n+1}}{1-r}.$

Conclusion. The formula holds for all $n\ge1$ .

Geometric sum by induction

Statement

Solution

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