$1+3+5^2+7+9^2+\ldots$ upto 40 terms is equal to

Question

Aicharya · Accepted Answer

$\left(1^2+5^2+9^2+\ldots \ldots 	ext { upto } 20 	ext { terms }ight)+(3+7+11+$...upto 20 terms) $\begin{aligned} & =\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-3)^2+\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-1) \ & =\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-3)^2+(4 \mathrm{r}-1) \ & =4 \sum_{\mathrm{r}=1}^{20}\left(4 \mathrm{r}^2-5 \mathrm{r}+2ight) \ & =16 \sum_{\mathrm{r}=1}^{20} \mathrm{r}^2-20 \sum_{\mathrm{r}=1}^{20} \mathrm{r}+8 \sum_{\mathrm{r}=1}^{20} 1=41880 \end{aligned}$ ~

$1+3+5^2+7+9^2+\ldots$ upto 40 terms is equal to

Solution