To Prove:

2 + 6 + 18 + … + 23^n–1 = (3ⁿ –1)

Steps to prove by mathematical induction:

Let P(n) be a statement involving the natural number n such that

(i) P(1) is true

(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ϵ N

Therefore,

Let P(n): 2 + 6 + 18 + … + 2 × 3^n–1 = (3ⁿ –1)

Step 1:

P(1) = 3¹ –1 = 3 - 1 = 2

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

P(k): 2 + 6 + 18 + … + 23^k–1 = (3^k –1)

Now,

2 + 6 + 18 + … + 2 × 3^k–1 + 2 × 3^{k + 1–1} = (3^k –1) + 2 × 3^k

= - 1 + 3 × 3^k

= 3^{k + 1 -} 1

= P(k + 1)

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

2 + 6 + 18 + … + 23^n–1 = (3ⁿ –1) for all n ϵ N