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Expert-verifiedFound in: Page 528

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 $n\ge 1$ **

**.**

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

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

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

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

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

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

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

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