Question

Let $S_n=1\cdot (n-1)+2\cdot (n-2)+3\cdot (n-3)+\dots +$ $(n-1)\cdot 1,n\ge 4$
The sum $\sum _{n=4}^{\infty }\left(\frac{2S_n}{n!}-\frac{1}{(n-2)!}\right)$ is equal to :

• A. $\frac{e-1}{3}$
• B. $\frac{e-2}{6}$
• C. $\frac{e}{3}$
• D. $\frac{e}{6}$

Answer 1

Answer: (A)

Let $T_r=r(n-r)$
$T_r=nr-r^2$
$\Rightarrow S_n=\sum _{r=1}^n{T}_r=\sum _{r=1}^n\left(nr-r^2\right)$
$S_n=\frac{n\cdot (n)(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}$
$S_n=\frac{n(n-1)(n+1)}{6}$
Now $\sum _{r=4}^{\infty }\left(\frac{2S_n}{n!}-\frac{1}{(n-2)!}\right)$
$=\sum _{r=4}^{\infty }\left(2\cdot \frac{n(n-1)(n+1)}{6\cdot n(n-1)(n-2)!}-\frac{1}{(n-2)!}\right)$
$=\sum _{r=4}^{\infty }\left(\frac{1}{3}\left(\frac{n-2+3}{(n-2)!}\right)-\frac{1}{(n-2)!}\right)$
$=\frac{1}{(n-3)!}=\frac{1}{3}(e-1)$