sum of series 1.2^2 + 3.3^2 + ..................n terms

Sum upto n^th term  (1.2² + 3.3² + ..................n terms)
Here the nth term is given by (2n-1) (n +1)²
So (2n-1) (n +1)²  = 2n³ +3n² -1
$$\begin{gathered} And \sum _{n=1}^{n=n}{n}^3 =\frac{\left\{n\right(n+1){\}}^2}{4} \\ And \sum _{n=1}^{n=n}{n}^2 =\frac{n(n+1)(2n+1)}{6} \\ So \sum _{n=1}^{n=n}(2n^3+3n^2-1) \\ =2 \times \frac{\left\{n\right(n+1){\}}^2}{4} +3\times \left\{\frac{n(n+1)(2n+1)}{6}\right\} -\left(1\right) Upto n times \\ =2 \times \frac{\left\{n\right(n+1){\}}^2}{4} +3\times \left\{\frac{n(n+1)(2n+1)}{6}\right\} -n \left(Ans\right) \end{gathered}$$