Thank you for reporting, we will resolve it shortly
Q.
The sum of the series $\frac{3}{1 !+2 !+3 !}+\frac{4}{2 !+3 !+4 !}+\frac{5}{3 !+4 !+5 !}+\ldots \ldots \ldots+\frac{2008}{(2006) !+(2007) !+(2008) !} \quad$ is equal to
Binomial Theorem
Solution:
$T _{ n }=\frac{ n }{( n -2) !+( n -1) !+ n !}=\frac{ n }{( n -2) ! \cdot[1+ n -1+ n ( n -1)]}=\frac{ n }{( n -2) ! \cdot n ^2}=\frac{1}{( n -2) ! \cdot n }$
$=\frac{n-1}{(n-1) ! \cdot n}=\frac{1}{(n-1) !}\left[1-\frac{1}{n}\right]=\frac{1}{(n-1) !}-\frac{1}{n !} $
$\text { nence } \operatorname{sum}=\displaystyle\sum_{n=3}^{2008}\left(\frac{1}{(n-1) !}-\frac{1}{n !}\right)$
sum $=\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{3 !}-1 / 4 !+1 / 4 !-1 / 5 ! \cdots \cdots \cdots+\frac{1}{(2007) !}-\frac{1}{(2008) !}=\frac{1}{2 !}-\frac{1}{(2008) !}=\frac{(2008) !-2}{2 \cdot(2008) !}$