Question

The sum of the series $\sum _{k=1}^{2011}\frac{k+2}{k!+(k+1)!+(k+2)!}$ is

• A. $\frac{1}{2}-\frac{1}{\left(2011\right)!}$
• B. $\frac{1}{2}-\frac{1}{\left(2012\right)!}$
• C. $\frac{1}{2}-\frac{1}{\left(2013\right)!}$
• D. $\frac{1}{2}-\frac{1}{\left(2014\right)!}$

Answer 1

The correct option is C $\frac{1}{2}-\frac{1}{\left(2013\right)!}$

Given,

$\begin{array}{c}{\sum }_{k=1}^{2011}\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}\\ {\sum }_{k=1}^{2011}\frac{k+2}{k!\left[k^2+4k+4\right]}={\sum }_{k=1}^{2011}\frac{k+2}{k!{\left(k+2\right)}^2}={\sum }_{k=1}^{2011}\frac{k+2}{k!\left(k+2\right)}\\ {\sum }_{k=1}^{2011}\frac{k+2}{k!\left(k+2\right)\left(k+1\right)}={\sum }_{k=1}^{2011}\frac{k+1}{|!k+2}\\ ={\sum }_{k=1}^{2011}\frac{k+2-1}{|!k+2}={\sum }_{k=1}^{2011}\left(\frac{1}{|!k+1}-\frac{1}{|!k+2}\right)\\ =\left(\frac{1}{|!2}-\frac{1}{|!3}\right)+\left(\frac{1}{|!3}-\frac{1}{|!4}\right)+...........+\left(\frac{1}{|!2012}-\frac{1}{|!2013}\right)\\ =\frac{1}{2}-\frac{1}{2013!}\end{array}$

Hence,

Option $C$ is correct answer.