Question

please answer this question​

Answer 1

Given Question :-

Prove that

$\rm \: a + ar + {ar}^{2} + - - - + {ar}^{n} = \dfrac{ {ar}^{n + 1} - a}{r - 1}$

To prove, this result, we use Principal of Mathematical Induction.

These are n + 1 number of terms,

So, Let assume that

$\rm :\longmapsto\:P(n + 1) : a + ar + {ar}^{2} +- - +{ar}^{n} = \dfrac{ {ar}^{n + 1} - a}{r - 1}$

Step : 1 For n = 1 ( means sum of first two terms )

$\rm :\longmapsto\:a + ar = \dfrac{ {ar}^{1 + 1} - a}{r - 1}$

$\rm :\longmapsto\:a + ar = \dfrac{ {ar}^{2} - a}{r - 1}$

$\rm :\longmapsto\:a + ar = \dfrac{ a({r}^{2} - 1)}{r - 1}$

$\rm :\longmapsto\:a + ar = \dfrac{ a(r - 1)(r + 1)}{r - 1}$

$\rm :\longmapsto\:a + ar = a(r + 1)$

$\rm :\longmapsto\:a + ar = ar +a$

$\bf\implies \:P(n) \: is \: true \: for \: n = 1$

Step : 2 Let assume that P(n+1) is true for n = k, where k is some natural number.

$\rm :\longmapsto\: a + ar + {ar}^{2} +- - +{ar}^{k} = \dfrac{ {ar}^{k + 1} - a}{r - 1}$

Step : 3 We have to prove that P(n+1) is true for n = k + 1 terms

$\rm :\longmapsto\:a + ar + {ar}^{2} +- - +{ar}^{k + 1} = \dfrac{ {ar}^{k + 2} - a}{r - 1}$

Consider, LHS

$\rm :\longmapsto\:a + ar + {ar}^{2} +- - +{ar}^{k + 1}$

$\rm \:  =  \: a + ar + {ar}^{2} +- - + {ar}^{k} +{ar}^{k + 1}$

$\rm \:  =  \: \dfrac{ {ar}^{k + 1} - a}{r - 1} +{ar}^{k + 1}$

$\rm \:  =  \: \dfrac{ {ar}^{k + 1} - a + {ar}^{k + 1} (r - 1)}{r - 1}$

$\rm \:  =  \: \dfrac{ {ar}^{k + 1} - a + {ar}^{k + 2} - {ar}^{k + 1} }{r - 1}$

$\rm \:  =  \: \dfrac{ - a + {ar}^{k + 2}}{r - 1}$

$\rm \:  =  \: \dfrac{{ar}^{k + 2} - a}{r - 1}$

Hence, By the Process of Principal of Mathematical Induction,

$\boxed{ \tt{ \: \rm \: a + ar + {ar}^{2} + - - - + {ar}^{n} = \dfrac{ {ar}^{n + 1} - a}{r - 1} \: }}$