A) \[\frac{n(n+3)}{4}\]
B) \[\frac{n(n+2)}{4}\]
C) \[\frac{n(n+1)(n+2)}{6}\]
D) \[{{n}^{2}}\]
E) \[0\]
Correct Answer: A
Solution :
\[\therefore \]\[{{T}_{n}}=\frac{1+2+3+....+n}{n}\] \[=\frac{\Sigma n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}\] Now, \[S=\Sigma {{T}_{n}}=\frac{1}{2}\Sigma n+\Sigma \frac{1}{2}\] \[=\frac{1}{2}\left\{ \frac{n(n+1)}{2} \right\}+\frac{n}{2}\] \[=\frac{n}{2}\left( \frac{n+1}{2}+1 \right)\] \[=\frac{n(n+3)}{4}\]
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