Prove by the method of induction, for all n ∈ N. 1^2 + 4^2 + 7^2 + …… + (3n – 2)^2 = n/2(6n^2 – 3n – 1)

Question

Prove by the method of induction, for all n ∈ N.

1² + 4² + 7² + …… + (3n – 2) = n/2(6n² – 3n – 1)

Answer 1

Let P(n) = 1² + 4² + 7² + ….. + (3n – 2)² = n/2

(6n² – 3n – 1), for all n ∈ N.

Step I:

Put n = 1

L.H.S.= 1² = 1

R.H.S.= 1/2[6(1)² – 3(1) – 1] = 1

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II:

Let us assume that P(n) is true for n = k.

∴ 1² + 4² + 7²+…..+ (3k – 2)² = k/2(6k – 3k – 1) ……(i)

Step III:

We have to prove that P(n) is true for n = k + 1

i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 1² + 4² + 7² + … + (3n – 2)² = n/2(6n² – 3n – 1) for all n ∈ N.