Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

Given,

n^th term of the series as:

a_n = (2n – 1)² = 4n² – 4n + 1

Then, the sum of n terms of the series can be expressed as

\(S_n= \displaystyle\sum_{k=1}^{n}a_k\)

The sum of first n terms of the given series is

\( S_n= \displaystyle\sum_{k=1}^{n}(4k^2-4k+1)\)

 \(= 4\displaystyle\sum_{k=1}^{n}k^2-4\displaystyle\sum_{k=1}^{n}k+4\displaystyle\sum_{k=1}^{n}1\)

= \( \dfrac{4n(n+1)(2n+1)}{6}-\)

= \( \dfrac{4n(n+1)}{2}+n\)

= \( \dfrac{2n(n+1)(2n+1)}{3}-2n(n+1)n\)

= \( n[ \dfrac{2(2n^2+3n+1)}{3}-2(n+1)n]\)

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

= \(n[ \dfrac{4n^2+1}{3}]\)

= \( \dfrac{n(2n+1)(2n-1)}{3}\)

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

\( \dfrac{n(2n+1)(2n-1)}{3}\)