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