Find the sum of nth term of this series and `S_n` denote the sum of its `n` terms. Then, `T_n=[1+(n-1xx2)^2]=(2n-1)^2=4n^2-4n+1`

Explanation

`T_n=(2n-1)^2`

`T_n=4n^n2-4n+1`

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

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

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

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

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