Maharashtra State BoardHSC Science (General) 11th
Advertisement Remove all ads

Prove by method of induction, for all n ∈ N: 12 + 32 + 52 + .... + (2n − 1)2 = n3(2n−1)(2n+1) - Mathematics and Statistics

Sum

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

12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`

Advertisement Remove all ads

Solution

Let P(n) ≡ 12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`, for all n ∈ N

Step I:

Put n = 1

L.H.S. = 12 = 1

R.H.S. = `1/3[2(1) - 1][(2(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 + 32 + 52 + .... + (2k − 1)2

= `"k"/3(2"k" - 1) (2"k" + 1)`   ...(i)

Step III:

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

i.e., to prove that

12 + 32 + 52 + …. + [2(k + 1) − 1]2

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

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

L.H.S. = 12 + 32 + 52 + …. + [2(k + 1) − 1]2 

= 12 + 32 + 52 + …. + (2k − 1)2 + (2k + 1)2

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

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

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

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

= `((2"k" + 1))/3[2"k"("k" + 1) + 3("k" + 1)]`

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

= 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 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`, for all n ∈ N.

Concept: Principle of Mathematical Induction
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 4 | Page 73
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×