The sum $1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots$ upto $\infty$ terms, is equal to

Question

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{S}=1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\ldots \ & =\sum_{\mathrm{r}=1}^{\infty} \frac{\mathrm{r}^2}{\mathrm{r}!} \ & =\sum_{\mathrm{r}=1}^{\infty} \frac{(\mathrm{r}-1+1)}{(\mathrm{r}-1)!}=\sum_{\mathrm{r}=2}^{\infty} \frac{1}{(\mathrm{r}-2)!}+\sum_{\mathrm{r}=1}^{\infty} \frac{1}{(\mathrm{r}-1)!} \ & =2 \mathrm{e}\end{aligned}$ /

The sum $1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots$ upto $\infty$ terms, is equal to

Solution