# Answer the following: Prove, by method of induction, for all n ∈ N 12 + 42 + 72 + ... + (3n − 2)2 = n2(6n2-3n-1) - Mathematics and Statistics

Sum

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

12 + 42 + 72 + ... + (3n − 2)2 = "n"/2 (6"n"^2 - 3"n" - 1)

#### Solution

Let P(n) ≡ 12 + 42 + 72 + .... + (3n − 2)2 = "n"/2 (6"n" − 3"n" − 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.

∴ P(n) is true for n = 1

Step II:

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

∴ 12 + 42 + 72 + .... + (3k − 2)2

= "k"/2(6"k"^2 - 3"k" - 1)   ...(i)

Step III:

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

i.e., to prove that

12 + 42 + 72 + …. + [3(k + 1) − 2]2

= (("k" + 1))/3[6("k" + 1)^2 - 3("k" + 1) - 1]

= (("k"  + 1))/2(6"k"^2 + 12"k" + 6 - 3"k" - 3 - 1)

= (("k" + 1))/2 (6"k"^2 + 9"k" + 2)

L.H.S. = 12 + 42 + 72 + …. + [3(k + 1) − 2]2

= 12 + 42 + 72 + …. + (3k − 2)2 + (3(k + 1) − 2]2

= "k"/2(6"k"^2 - 3"k" - 1) + (3"k" + 1)^2   ...[From (i)]

= ((6"k"^3 - 3"k"^2 - "k") + 2(9"k"^2 + 6"k" + 1))/2

= (6"k"^3 + 15"k"^2 + 11"k" + 2)/2

= (("k" + 1)(6"k"^2 + 9"k" + 2))/2

= R.H.S.

∴ P(n) is true for n = k + 1

Step IV:

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

∴ 12 + 42 + 72 + ... + (3n − 2)2 = "n"/2 (6"n"^2 - 3"n" - 1) for all n ∈ N

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4 | Q II. (1) (ii) | Page 85