Question

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

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Short answer

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

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

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" $$\begin{gathered} Sr=r\left(1+r+r^2+r^3+....r^{n-1}\right) \\ Sr=r+r^2+r^3+.....+r^{n-1}+r^n.....\left(2\right) \end{gathered}$$

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