# Prove the following by using the principle of mathematical induction for all n ∈ N:

1.3 + 3.5 + 5.7 + ..... + (2n - 1)(2n + 1) = [n (4n² + 6n - 1)]/3

**Solution:**

Let P (n) be the given statement

i.e., P (n) : 1.3 + 3.5 + 5.7 + ..... + (2n - 1)(2n + 1) = [n (4n² + 6n - 1)]/3

P (1) :1.3 = [1 (4.1² + 6.1 - 1)]/3

3 = 9/3

3 = 3, which is true.

Assume that P (k) is true for some positive integer k.

i.e., P (n) : 1.3 + 3.5 + 5.7 + ..... + (2k - 1)(2k + 1) = [k (4k² + 6k - 1)]/3 ....(1)

We will now prove that P (k + 1) is also true.

Now, we have

LHS = 1.3 + 3.5 + 5.7 + ..... + [2 (k + 1) - 1] [2 (k + 1) + 1]

= [1.3 + 3.5 + 5.7 + ..... + (2k - 1)(2k + 1)] + (2k + 1)(2k + 3)

= [k (4k² + 6k - 1)]/3 + (4k² + 8k + 3) .... [from (1)]

= [k (4k² + 6k - 1) + 3(4k² + 8k + 3)]/3

= (4k³ + 6k² - k + 12k² + 24k + 9)/3

= (4k³ + 18k² + 23k + 9)/3 ... (2)

RHS = (k + 1) [4 (k + 1)² + 6 (k + 1) - 1]/3

= (k + 1) [4 (k² + 2k + 1) + 6 (k + 1) - 1]/3

= (k + 1)(4k² + 8k + 4 + 6k + 6 - 1)/3

= (k + 1)(4k² + 14k + 9)/3

= (4k³ + 18k² + 23k + 9)/3 ... (3)

From (2) and (3), LHS = RHS.

Thus P (k + 1) is true, whenever P (k) is true.

Hence, from the principle of mathematical induction, the statement P (n) is true for all natural numbers i.e., n ∈ N .

NCERT Solutions Class 11 Maths Chapter 4 Exercise 4.1 Question 7

## Prove the following by using the principle of mathematical induction for all n ∈ N: 1.3 + 3.5 + 5.7 + ..... + (2n - 1)(2n + 1) = [n (4n² + 6n - 1)]/3

**Summary:**

We have proved that 1.3 + 3.5 + 5.7 + ..... + (2n - 1)(2n + 1) = [n (4n² + 6n - 1)]/3 by using the principle of mathematical induction for all n ∈ N.

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