A) \[\frac{1}{n}\]
B) \[\frac{1}{n+1}\]
C) \[\frac{2\,\,(n-1)}{n}\]
D) \[\frac{n}{n+1}\]
Correct Answer: D
Solution :
\[{{T}_{n}}=\frac{1}{n\,\,(n+1)}=\left( \frac{1}{n}-\frac{1}{n+1} \right)\] \[\therefore \] Given sum\[=\left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\] \[\left( \frac{1}{4}-\frac{1}{5} \right)+...+\left\{ \frac{1}{n}-\frac{1}{(n+1)} \right\}=\left\{ 1-\frac{1}{n+1} \right\}=\frac{n}{(n+1)}\]
You need to login to perform this action.
You will be redirected in
3 sec
