# Solve the Following Questions

Question:

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 :

1. $\frac{\mathrm{e}-1}{3}$

2. $\frac{\mathrm{e}-2}{6}$

3. $\frac{\mathrm{e}}{3}$

4. $\frac{\mathrm{e}}{6}$

Correct Option: 1

Solution:

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)

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