Find the sum to n terms of the series, whose `n^(t h)`terms is given by : `(2n-1)^2`

Explanation

Here, `a_n = (2n-1)^2 = 4n^2+1-4n`

So, `S_n = sum_(n=1)^na_n = sum_(n=1)^n4n^2+sum_(n=1)^n1-sum_(n=1)^n4n`

`S_n = 4((n(n+1)(2n+1))/6)+n-4((n(n+1))/2)`

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

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

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

`=4/3n(n+1)(n-1)+n`

`=n(4/3(n^2-1)+1)`

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