Question

Using the principle of mathematical induction, prove each of the following for all n ϵ N: 

2 + 6 + 18 + … + 2 × 3^n–1 = (3ⁿ –1)

Answer 1

To Prove:

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