Question

If in a series $t_n=\frac{n+1}{(n+2)!}$ then $\sum _{n=0}^{10}{t}_n$ is equal to?

• A. $1-\frac{1}{10!}$
• B. $1-\frac{1}{11!}$
• C. $1-\frac{1}{12!}$
• D. None of these

Answer 1

The correct option is C $1-\frac{1}{12!}$

$\begin{array}{c}\text{Solution - It is given that}\\ \begin{array}{cc}{t}_n& =\frac{n+1}{(n+2)!}\\ tn& =\frac{n+1+1-1}{(n+2)!}=\frac{(n+2)-1}{(n+2)!}\\ & \text{th}=\frac{n+2}{(n+2)!}-\frac{1}{(n+2)!}\end{array}\end{array}$

$tn=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$ ${\sum }_{n=0}^{10}{t}_n={\sum }_{n=0}^{10}(\frac{1}{(n+1)!}-\frac{1}{(n+2)!})$

$\begin{array}{cc}& =\frac{1}{1}(-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}\cdots \cdots -\frac{1}{11!}+\frac{1}{11!})-\frac{1}{12!}\\ & =1-\frac{1}{12!}\\ \text{Hence,}& \left(C\right)\text{is the correct option.}\end{array}$