• # question_answer The sum $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...+\frac{k+2}{k!+(k+1)!+(k+2)!}$ $+...+\frac{2009}{(2007)!\,+\,(2008)!\,+\,(2009)!}$ A)  $\frac{1}{3!}+\frac{1}{(2010)!}$               B)  $\frac{1}{2}+\frac{1}{(2010)!}$ C) $\frac{1}{2}-\frac{1}{(2009)!}$                   D)  $\frac{1}{3!}-\frac{1}{(2009)!}$

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 $={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+\,{{T}_{2007}}$ $=\left[ \left( \frac{1}{2!}-\frac{1}{3!} \right)+\left( \frac{1}{3!}-\frac{1}{4!} \right) \right.$ $\left. +...+\frac{1}{(2008)!}-\frac{1}{(2009)!} \right]$ $=\frac{1}{2}-\frac{1}{(2009)!}$

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