# Prove the following by using the principle of mathematical induction for all n ∈ N:

Question:

Prove the following by using the principle of mathematical induction for all $n \in N: a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a\left(r^{n}-1\right)}{r-1}$

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a\left(r^{n}-1\right)}{r-1}$

For $n=1$, we have

$\mathrm{P}(1): a=\frac{a\left(r^{1}-1\right)}{(r-1)}=a$, which is true.

Let P(k) be true for some positive integer k, i.e.,

$a+a r+a r^{2}+\ldots \ldots+a r^{k-1}=\frac{a\left(r^{k}-1\right)}{r-1}$  $\ldots$ (i)

We shall now prove that P(k + 1) is true.

Consider

$\left\{a+a r+a r^{2}+\ldots \ldots+a r^{k-1}\right\}+a r^{(k+1)-1}$

$=\frac{a\left(r^{k}-1\right)}{r-1}+a r^{k}$ $[$ Using $(i)]$

$=\frac{a\left(r^{k}-1\right)+a r^{k}(r-1)}{r-1}$

$=\frac{a\left(r^{k}-1\right)+a r^{k+1}-a r^{k}}{r-1}$

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

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

$=\frac{a\left(r^{k+1}-1\right)}{r-1}$

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.