Question

The sum of the series $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\dots \dots \dots +\frac{2008}{(2006)!+(2007)!+(2008)!}$ is equal to

• A. $\frac{(2008)!+2}{2\cdot (2008)!}$
• B. $\frac{(2008)!+1}{2\cdot (2008)!}$
• C. $\frac{(2008)!-2}{2\cdot (2008)!}$
• D. $\frac{(2008)!-3}{2\cdot (2008)!}$

Answer 1

Answer: (C)

$T_n=\frac{n}{(n-2)!+(n-1)!+n!}=\frac{n}{(n-2)!\cdot [1+n-1+n(n-1)]}=\frac{n}{(n-2)!\cdot n^2}=\frac{1}{(n-2)!\cdot n}$
$=\frac{n-1}{(n-1)!\cdot n}=\frac{1}{(n-1)!}\left[1-\frac{1}{n}\right]=\frac{1}{(n-1)!}-\frac{1}{n!}$
$\text{ nence }\mathrm{sum}=\sum _{n=3}^{2008}\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)$
sum $=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-1/4!+1/4!-1/5!\cdots \cdots \cdots +\frac{1}{(2007)!}-\frac{1}{(2008)!}=\frac{1}{2!}-\frac{1}{(2008)!}=\frac{(2008)!-2}{2\cdot (2008)!}$