A) \[\frac{7}{2}\]
B) \[\frac{11}{4}\]
C) \[\frac{11}{2}\]
D) \[\frac{60}{11}\]
Correct Answer: C
Solution :
Given sum is \[\frac{3}{12}+\frac{5}{{{1}^{2}}+{{2}^{2}}}+\frac{7}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.....\] \[nth\,\,term={{T}_{n}}\] \[=\frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}=\frac{6}{n(n+1)}\] or \[{{T}_{n}}=6\left[ \frac{1}{n}-\frac{1}{n+1} \right]\] \[\therefore \] \[{{S}_{n}}=\] \[\sum{{{T}_{n}}=6}\sum{\frac{1}{n}-6}\sum{\frac{1}{n+1}}=\frac{6n}{n}-\frac{6}{n+1}\] \[=6-\frac{6}{n+1}=\frac{6n}{n+1}\] So, sum upto 11 terms means \[{{S}_{11}}=\frac{6\times 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
