A) \[\frac{4n}{(n+1)}\]
B) \[\frac{6n}{(n+1)}\]
C) \[\frac{12n}{(n+1)}\]
D) \[\frac{9n}{(n+1)}\]
Correct Answer: B
Solution :
\[\frac{3}{{{1}^{2}}}+\frac{5}{{{1}^{2}}+{{2}^{2}}}+\frac{7}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.....n\]terms \[{{T}_{n}}=\frac{2n+1}{\Sigma {{n}^{2}}}\] \[=\frac{2n+1}{\left( n\frac{(2n+1)(n+1)}{6} \right)}\] \[=\left( \frac{6}{n(n+1)} \right)\] \[=6\left[ \frac{1}{n}-\frac{1}{(n+1)} \right]\] \[\therefore \] \[{{S}_{n}}=6\left[ 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4} \right.+\] \[\left. .....+\frac{1}{n}-\frac{1}{n+1} \right]\] \[=6\left[ 1-\frac{1}{n+1} \right]\] \[=6\left( \frac{n+1-1}{n+1} \right)=\frac{6n}{n+1}\]
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