Mathematics - Sequence & Series Question with Solution | TestHub

$\left(1^2+5^2+9^2+\dots \dots \text{ upto }20\text{ terms }\right)+(3+7+11+$...upto 20 terms)
$\begin{array}{rl}& =\sum _{\mathrm{r}=1}^{20}(4\mathrm{r}-3{)}^2+\sum _{\mathrm{r}=1}^{20}(4\mathrm{r}-1)\\ & =\sum _{\mathrm{r}=1}^{20}(4\mathrm{r}-3{)}^2+(4\mathrm{r}-1)\\ & =4\sum _{\mathrm{r}=1}^{20}\left(4{\mathrm{r}}^2-5\mathrm{r}+2\right)\\ & =16\sum _{\mathrm{r}=1}^{20}{\mathrm{r}}^2-20\sum _{\mathrm{r}=1}^{20}\mathrm{r}+8\sum _{\mathrm{r}=1}^{20}1=41880\end{array}$