The sum of first n terms of the series $1^2+\left(1\right)\left(2\right)+3^2+\left(3\right)\left(4\right)+5^2+\left(5\right)\left(6\right)+7^2+\dots \dots$ when n is odd is

$\begin{array}{rcl} & & 1^2+1\left(2\right)+3^2+3\left(4\right)+5^2+5\left(6\right)+......... (\mathrm{n} \mathrm{is} \mathrm{odd})\\ & & 1^2+2(2-1)+3^2+4(4-1)+5^2+6(6-1)+......+(\mathrm{n}-1) (\mathrm{n}-1-1)+{\mathrm{n}}^2\\ & =& 1^2+2^2+3^2+........+{\mathrm{n}}^2-\left(2+4+6+.....+(\mathrm{n}-1)\right)\\ & =& \frac{\mathrm{n}(\mathrm{n}+1) (2\mathrm{n}+1)}{6}-\frac{2{\displaystyle \frac{\mathrm{n}-1}{2}} \left({\displaystyle \frac{\mathrm{n}-1}{2}}+1\right)}{2}\\ & =& \frac{\mathrm{n}(\mathrm{n}+1) (2\mathrm{n}+1)}{6}-\frac{(\mathrm{n}-1) (\mathrm{n}+1)}{4}\\ & =& \frac{(\mathrm{n}+1)}{1} \left[\frac{\mathrm{n}(2\mathrm{n}+1)}{3}-\frac{\mathrm{n}-1}{2}\right]\\ & =& \frac{\mathrm{n}+1}{2} \left[\frac{4{\mathrm{n}}^2+2\mathrm{n}-3\mathrm{n}+3}{6}\right]\\ & =& \frac{(\mathrm{n}+1) (4{\mathrm{n}}^2-\mathrm{n}+3)}{12}\end{array}$