A) \[\frac{1}{3!}+\frac{1}{(2010)!}\]
B) \[\frac{1}{2}+\frac{1}{(2010)!}\]
C) \[\frac{1}{2}-\frac{1}{(2009)!}\]
D) \[\frac{1}{3!}-\frac{1}{(2009)!}\]
Correct Answer: C
Solution :
We have, \[{{T}_{n}}=\frac{n+2}{n!\,+(n+1)!+(n+2)!}\] \[=\frac{n+2}{n!\,(1+n+1+{{n}^{2}}+3n+2)}\] \[=\frac{n+2}{n!({{n}^{2}}+4n+4)}=\frac{n+2}{n!\,{{(n+2)}^{2}}}\] \[=\frac{1}{(n+2)\,n!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}\] Sum \[={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+\,{{T}_{2007}}\] \[=\left[ \left( \frac{1}{2!}-\frac{1}{3!} \right)+\left( \frac{1}{3!}-\frac{1}{4!} \right) \right.\] \[\left. +...+\frac{1}{(2008)!}-\frac{1}{(2009)!} \right]\] \[=\frac{1}{2}-\frac{1}{(2009)!}\]
You need to login to perform this action.
You will be redirected in
3 sec
