Sum to 20 terms of the series $1{.3}^2+2{.5}^2+3{.7}^2+\dots$ is

We have,

${\mathrm{t}}_{\mathrm{n}} = [\mathrm{nth} \mathrm{term} \mathrm{of} 1, 2, 3, \dots ] \times {[\mathrm{nth} \mathrm{term} \mathrm{of} 3, 5, 7, \dots ]}^2$

$=\mathrm{n}(2\mathrm{n}+1{)}^2=4{\mathrm{n}}^3+4{\mathrm{n}}^2+\mathrm{n}$

$\begin{array}{l}\therefore {\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{t}}_{\mathrm{n}}=4\sum {\mathrm{n}}^3+4\sum {\mathrm{n}}^2+\sum \mathrm{n}\\ =4\cdot {\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]}^2+4\cdot \frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}\\ ={\mathrm{n}}^2(\mathrm{n}+1{)}^2+\frac{2}{3}\mathrm{n}(\mathrm{n}+1\left)\right(2\mathrm{n}+1)+\frac{1}{2}\cdot \mathrm{n}(\mathrm{n}+1)\\ \therefore {\mathrm{S}}_{20}={20}^2\cdot {21}^2+\frac{2}{3}\times 20\cdot 21\cdot 41+\frac{1}{2}20 \left(21\right)=188090\\ \end{array}$