A) \[\frac{n-1}{n}\]
B) \[\frac{n}{n+1}\]
C) \[\frac{n+1}{n+2}\]
D) \[\frac{n+1}{n}\]
Correct Answer: B
Solution :
[b] The general term is \[{{T}_{n}}=\frac{\frac{n}{2}.\frac{n+1}{2}}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+....+{{n}^{3}}}=\frac{1}{n(n+1)}\] \[=\frac{1}{n}-\frac{1}{n+1}\] \[\therefore \,\,\,\,\,\,\,\,{{S}_{n}}=1-\frac{1}{n+1}=\frac{n}{n+1}\]
You need to login to perform this action.
You will be redirected in
3 sec
