A) \[\frac{2n}{n+1}\]
B) \[\frac{4n}{n+1}\]
C) \[\frac{6n}{n+1}\]
D) \[\frac{9n}{n+1}\]
Correct Answer: C
Solution :
\[{{T}_{n}}=\frac{(2n+1)}{n(n+1)(2n+1)/6}=\frac{6}{n(n+1)}\] \[{{S}_{n}}=\Sigma ({{T}_{n}})=\Sigma \,6\,\left[ \frac{1}{n}-\frac{1}{n+1} \right]\]\[=6\left[ 1-\frac{1}{n+1} \right]\] \[{{S}_{n}}=\frac{6n}{n+1}\].
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