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

a + ar + ar² + .... + arⁿ ⁻ ¹ = a (rⁿ - 1)/(r - 1)

**Solution:**

Let P (n) be the given statement.

i.e., P (n) : a + ar + ar² + .... + ar^{n - 1} = a (r^{n} - 1)/(r - 1)

For n = 1,

P (1) : a = a (r^{1} - 1)/(r - 1)

a = a (r - 1)/(r - 1)

a = a , which is true.

Assume that P (k) is true for some positive integer k.

i.e., a + ar + ar² + .... + ar^{k - 1} = a (r^{k} - 1)/(r - 1) ....(1)

a + ar + ar² + .... + ar^{(k + 1) - 1}

= [a + ar + ar ^{2} + ..... + ar^{k }^{- 1}] + ar^{k}

= a (r^{k} - 1)/(r - 1) + ar^{k} [from(1)]

= [a (r^{k} - 1) + ar^{k} (r - 1)]/(r - 1)

= [ar^{k} - a + ar^{k }^{+ 1} - ar^{k}] / (r - 1)

= (ar^{k }^{+ 1} - a)/(r - 1)

= a (r^{k }^{+ 1} - 1)/(r - 1)

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

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

NCERT Solutions Class 11 Maths Chapter 4 Exercise 4.1 Question 12

## Prove the following by using the principle of mathematical induction for all n ∈ N: a + ar + ar² + .... + arⁿ ⁻ ¹ = a (rⁿ - 1)/(r - 1)

**Summary:**

We have proved that a + ar + ar² + .... + arⁿ ⁻ ¹ = a (rⁿ - 1)/(r - 1) by using the principle of mathematical induction for all n ∈ N.