Answer :

To Prove:


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


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