The sum of first n terms of the series $1^2+2{.2}^2+3^2+2{.4}^2+5^2+5{.6}^2+\dots \mathrm{is} \frac{\mathrm{n}(\mathrm{n}+1{)}^2}{2}$when n is even. When n is odd the sum is

When n is odd, last term will be n² , $\therefore$ then the sum is

$$\begin{gathered} 1^2+2{.2}^2+3^2+2{.4}^2+5^2+2{.6}^2+\dots +2(\mathrm{n}-1{)}^2+{\mathrm{n}}^2 \\ \begin{array}{l}=\frac{(\mathrm{n}-1){\mathrm{n}}^2}{2}+{\mathrm{n}}^2\left[\text{ Replacing }\mathrm{n}\text{ by }\mathrm{n}-1\text{ in }\frac{\mathrm{n}(\mathrm{n}+1{)}^2}{2}\right]\\ =\frac{{\mathrm{n}}^3-{\mathrm{n}}^2+2{\mathrm{n}}^2}{2}=\frac{{\mathrm{n}}^3+{\mathrm{n}}^2}{2}=\frac{{\mathrm{n}}^2(\mathrm{n}+1)}{2}\end{array} \end{gathered}$$