a+ar+ar*2+....+ar*n-1=a(r*n-1)/r-1

Let P(n) = a + ar + ar² + ..... + ar^n-1{tex} = \frac{{a({r^n} - 1)}}{{r - 1}}{/tex}.
For n = 1
{tex}P(1) = a{r^{1 - 1}} = \frac{{a({r^1} - 1)}}{{r - 1}} \Rightarrow a = a{/tex}
{tex}\therefore {/tex} P(1) is true
Let P(n) be true for n = k
{tex}\therefore P(k) = a + ar + a{r^2} + .... + a{r^{k - 1}}{/tex}{tex} = \frac{{a({r^k} - 1)}}{{r - 1}}{/tex} .... (i)
For n = k + 1
R.H.S. {tex} = \frac{{a({r^{k + 1}} - 1)}}{{r - 1}}{/tex}
L.H.S. {tex} = \frac{{a({r^k} - 1)}}{{r - 1}} + a{r^k}{/tex} [ Using (i)]
{tex} = \frac{{a{r^k}}}{{r - 1}} - \frac{a}{{r - 1}} + a{r^k}{/tex}
{tex} = a{r^k} \cdot \left( {\frac{1}{{r - 1}}+1} \right) - \frac{a}{{r - 1}} = \frac{{a{r^{k + 1}}}}{{r - 1}} - \frac{a}{{r - 1}}{/tex}{tex} = \frac{{a{r^{k + 1}} - a}}{{r - 1}}{/tex}
{tex} = \frac{{a({r^{k + 1}} - 1)}}{{r - 1}}{/tex}
{tex}\therefore {/tex} P(k + 1) is true
Thus P(k) is true {tex} \Rightarrow {/tex} P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all {tex}n \in N{/tex}.