A) \[\frac{n-1}{2}\]
B) \[\frac{n}{n+1}\]
C) \[\frac{n+1}{n+2}\]
D) \[\frac{(n+1)}{n}\]
Correct Answer: B
Solution :
\[{{T}_{n}}=\frac{\frac{n(n+1)}{2.2}}{\sum {{n}^{3}}}=\frac{\frac{n(n+1)}{4}}{\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}}\] \[=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\] \[\therefore \] \[{{S}_{n}}=\sum \left( \frac{1}{n}-\frac{1}{n+1} \right)\] \[=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\] \[=1-\frac{1}{n+1}=\frac{n}{n+1}\]You need to login to perform this action.
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