Find the sum to n terms of the series whose n^th term is given by (2n - 1)²

Solution:

The given n^th term is a_n = (2n - 1)²

Hence,

a_n = (2n - 1)²

= 4n² - 4n + 1

Therefore,

S_n = ∑ⁿ_{k = 1}(a)_k

S_n = ∑ⁿ_{k = 1}(4k² - 4k + 1)

= 4∑ⁿ_{k = 1}(k)² + 4∑ⁿ_{k = 1}(k) + ∑ⁿ_{k = 1}(1)

= [4n (n + 1)(2n + 1)]/6 - [4n (n + 1)]/2 + n

= [2n (n + 1)(2n + 1)]/3 - 2n (n + 1) + n

= n [2 (2n² + 3n + 1)/3 - 2 (n + 1) + 1]

= n [(4n² + 6n + 2 - 6n - 6 + 3)/3]

= n [(4n² - 1)/3]

= n/3 (2n + 1)(2n - 1)

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NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.4 Question 10

Find the sum to n terms of the series whose n^th term is given by (2n - 1)²

Summary:

We knew that a_n = (2n - 1)² and therefore S_n = ∑ⁿ_{k = 1}(a)_k so the sum to n terms of the series is n/3 (2n + 1)(2n - 1)