Question

a + ar + ar² + ... + arⁿ⁻¹ = $a\left(\frac{r^n-1}{r-1}\right),r\ne 1$

Answer 1

Let P(n) be the given statement.
Now,
$$\begin{gathered} P\left(n\right)=a+ar+a{r}^2+...+a{r}^{n-1}=a\left(\frac{r^n-1}{r-1}\right),r\ne 1 \\ \mathrm{Step}1: \\ P\left(1\right)=a=a\left(\frac{r^1-1}{r-1}\right) \\ \mathrm{Hence},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Suppose}P\left(m\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Then}, \\ a+ar+a{r}^2+...+a{r}^{m-1}=a\left(\frac{r^m-1}{r-1}\right),r\ne 1 \\ Toshow:P(m+1)\mathrm{is}\mathrm{true}\mathrm{whenever}P\left(m\right)\mathrm{is}\mathrm{true}. \\ Thatis, \\ a+ar+a{r}^2+...+a{r}^m=a\left(\frac{r^{m+1}-1}{r-1}\right),r\ne 1 \\ We\mathrm{know}\mathrm{that}P\left(m\right)\mathrm{is}\mathrm{true}. \\ Thus,wehave: \\ a+ar+a{r}^2+...+a{r}^{m-1}=a\left(\frac{r^m-1}{r-1}\right) \\ \Rightarrow a+ar+a{r}^2+...+a{r}^{m-1}+a{r}^m=a\left(\frac{r^m-1}{r-1}\right)+a{r}^m\left[\mathrm{Adding}a{r}^m\mathrm{to}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow P(m+1)=a\left(\frac{r^m-1+r.r^m-r^m}{r-1}\right) \\ \Rightarrow P(m+1)=a\left(\frac{r^{m+1}-1}{r-1}\right),r\ne 1 \\ \mathrm{Thus},P(m+1)istrue. \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N. \end{gathered}$$