A) \[\frac{n(n+3)}{4}\]
B) \[\frac{n(n+2)}{4}\]
C) \[\frac{n(n+1)\,(n+2)}{6}\]
D) \[{{n}^{2}}\]
Correct Answer: A
Solution :
\[{{T}_{n}}=\frac{1+2+3+.....+n}{n}=\frac{n(n+1)}{2n}=\frac{1}{2}(n+1)\] Hence, \[S=\frac{1}{2}(\Sigma n+n)\]\[=\frac{1}{2}\left\{ \frac{n(n+1)}{2}+n \right\}=\frac{n(n+3)}{4}\].
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