If n is even, then the sum to first n terms of the series $1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+\dots \dots$ is

$\begin{array}{rcl} & & 1^2+2·2^2+3^2+2·4^2+......+2·{\mathrm{n}}^2 (\mathrm{n} \mathrm{is} \mathrm{even})\\ & =& 1^2+2^2+3^2+........+{\mathrm{n}}^2+\left(2^2+4^2+......+{\mathrm{n}}^2\right)\\ & =& \frac{\mathrm{n}(\mathrm{n}+1) (2\mathrm{n}+1)}{6}+\frac{4 {\displaystyle \frac{\mathrm{n}}{2}} \left({\displaystyle \frac{\mathrm{n}}{2}}+1\right) (\mathrm{n}+1)}{6}\\ & =& \frac{\mathrm{n}(\mathrm{n}+1)}{6} (2\mathrm{n}+1+\mathrm{n}+2)\\ & =& \frac{\mathrm{n}(\mathrm{n}+1)}{6} (3\mathrm{n}+3)\\ & =& \frac{\mathrm{n}{(\mathrm{n}+1)}^2}{2}\end{array}$