Answer : To Prove:

2 + 6 + 18 + … + 2  3n–1 = (3n –1)

Steps to prove by mathematical induction:

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

• P(1) is true
• 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 × 3n–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 + … + 2  3k–1 = (3k –1)

Now,

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

= – 1 + 3 × 3k

= 3k + 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 + … + 2       3n–1 = (3n –1) for all n ϵ N