Suggested languages for you:

Americas

Europe

4

Expert-verified
Found in: Page 528

### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

# Let r be a real number, $r\ne 1$. Prove that for every integer $n\ge 1$ , $1+r+{r}^{2}+{r}^{3}+....+{r}^{n-1}=\frac{{r}^{n}-1}{r-1}$.

It is proved that for every integer $n\ge 1$:

$1+r+{r}^{2}+{r}^{3}+....+{r}^{n-1}=\frac{{r}^{n}-1}{r-1}$

See the step by step solution

## Consider the given parameters

Assume that $r\ne 1$ is a real number. The objective is to show that for each integer $n\ge 1$ .

$1+r+{r}^{2}+{r}^{3}+....+{r}^{n-1}=\frac{{r}^{n}-1}{r-1}$

## Obtain the equation for Sr

Let $S=1+r+{r}^{2}+{r}^{3}+...+{r}^{n-1}$ …. (1)

Then multiply both the sides of the sum S by the real number r and get the equation as follows:

role="math" localid="1648726203258" $Sr=r\left(1+r+{r}^{2}+{r}^{3}+....{r}^{n-1}\right)\phantom{\rule{0ex}{0ex}}Sr=r+{r}^{2}+{r}^{3}+.....+{r}^{n-1}+{r}^{n}.....\left(2\right)$

## Solve the equations (1) and (2)

Subtract equation (1) from equation (2) and solve in the following manner:

$\begin{array}{rcl}rS-S& =& r+{r}^{2}+{r}^{3}+....+{r}^{n-1}+{r}^{n}-\left(1+r+{r}^{2}+{r}^{3}+....{r}^{n-1}\right)\\ \left(r-1\right)S& =& r+{r}^{2}+{r}^{3}+....+{r}^{n-1}+{r}^{n}-1-r-{r}^{2}-{r}^{3}-.....-{r}^{n-1}\\ \left(r-1\right)S& =& {r}^{n}-1\\ S& =& \frac{{r}^{n}-1}{r-1}\end{array}$

$S=1+r+{r}^{2}+{r}^{3}+....+{r}^{n-1}$; therefore, $1+r+{r}^{2}+{r}^{3}+.......+{r}^{n-1}=\frac{{r}^{n}-1}{r-1}$ .