Question

By the principle of mathematical induction, prove the following:

1 + 4 + 7 + ……. + (3n – 2) = `("n"(3"n" - 1))/2`  for all n ∈ N.

Answer 1

Let P(n) : 1 + 4 + 7 + ……. + (3n – 2) = `("n"(3"n" - 1))/2`

Put n = 1,

LHS = 1

RHS = `(1(3 - 1))/2 = 1`

∴ P(1) is true.

Assume P(k) is true for n = k

P(k): 1 + 4 + 7 + ……. + (3k – 2) = `(k(3k - 1))/2`

To prove P(k + 1) is true, i.e., to prove

1 + 4 + 7 + ……. + (3k – 2) + (3(k + 1) – 2) = `((k + 1)(3(k + 1) - 1))/2`

1 + 4 + 7 + ……. + (3k – 2) + (3k + 3 – 2) = `((k + 1)(3k + 2))/2`

1 + 4 + 7 + …… + (3k + 1) = `((k + 1)(3k + 2))/2`

1 + 4 + 7 + …… + (3k – 2) + (3k + 1) = `(k(3k - 1))/2 + (3k + 1)`

`= (k(3k - 1) + 2(3k + 1))/2`

`= (3k^2 - k + 6k + 2)/2`

`= (3k^2 + 5k + 2)/2`

`= ((k + 1)(3k + 2))/2`

∴ P(k + 1) is true whenever P(k) is true.

∴ By the Principle of Mathematical Induction, P(n) is true for all n ∈ N.