# 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}

= 4n^{2} - 4n + 1

Therefore,

S_{n} = ∑^{n}_{k = 1}(a)_{k}

S_{n} = ∑^{n}_{k = 1}(4k^{2} - 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 (2n^{2} + 3n + 1)/3 - 2 (n + 1) + 1]

= n [(4n^{2} + 6n + 2 - 6n - 6 + 3)/3]

= n [(4n^{2} - 1)/3]

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

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)