Let $S_{n}=1 \cdot(n-1)+2 \cdot(n-2)+3 \cdot(n-3)+\ldots+$ $(\mathrm{n}-1) \cdot 1, \mathrm{n} \geq 4$
The sum $\sum_{n=4}^{\infty}\left(\frac{2 S_{n}}{n !}-\frac{1}{(n-2) !}\right)$ is equal to :
Correct Option: 1
Let $\mathrm{T}_{\mathrm{r}}=\mathrm{r}(\mathrm{n}-\mathrm{r})$
$\mathrm{T}_{\mathrm{r}}=\mathrm{nr}-\mathrm{r}^{2}$
$\Rightarrow S_{n}=\sum_{r=1}^{n} T_{r}=\sum_{r=1}^{n}\left(n r-r^{2}\right)$
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n} .(\mathrm{n})(\mathrm{n}+1)}{2}-\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}+1)}{6}$
Now $\sum_{r=4}^{\infty}\left(\frac{2 S_{n}}{n !}-\frac{1}{(n-2) !}\right)$
$=\sum_{\mathrm{r}=4}^{\infty}\left(2 \cdot \frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}+1)}{6 \cdot \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}-\frac{1}{(\mathrm{n}-2) !}\right)$
$=\sum_{\mathrm{r}=4}^{\infty}\left(\frac{1}{3}\left(\frac{\mathrm{n}-2+3}{(\mathrm{n}-2) !}\right)-\frac{1}{(\mathrm{n}-2) !}\right)$
$=\sum_{\mathrm{r}=4}^{\infty} \frac{1}{3} \cdot \frac{1}{(\mathrm{n}-3) !}=\frac{1}{3}(\mathrm{e}-1)$
Option (1)
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.