Question

Use mathematical induction to prove that if $r\ne 1$then

role="math" localid="1647843083263" $a+ar+a{r}^2+...+a{r}^{n-1}=a\left(\frac{1-r^n}{1-r}\right)$

Short answer

The given statement is shown.

Step-by-step solution

Step 1. Given information

$r\ne 1$

Step 2. Prove that a+ar+ar2+...+arn-1=a1-rn1-r.

First we consider the above statement for $n=1$

Since $a=a\frac{1-r}{1-r}$, we say that above statement for $n=1$

Hence condition holds true.

Now we assume that above statement is true for some $k$

So we have,

$a+ar+a{r}^2+\cdots +a{r}^{k-1}=a\frac{1-r^k}{1-r}$

Now let us consider the sum of first $k+1$terms

$\begin{array}{rcl}a+ar+a{r}^2+\cdots +a{r}^{k-1}+a{r}^{k+1-1}& =& \left(a+ar+a{r}^2+\cdots +a{r}^{k-1}\right)+a{r}^k\\ & =& a\frac{1-r^k}{1-r}+a{r}^k\\ & =& a\frac{1-r^{k+1}}{1-r}\end{array}$

This shows that the statement is for $n=k+1$

Hence condition holds true.

As a result, we say that given statement is true for any natural number$n$