A) \[\frac{n(n+1)(n+2)}{3}\]
B) \[\frac{(n+1)(n+2)(n+3)}{12}\]
C) \[{{n}^{2}}(n+2)\]
D) \[n(n+1)(n+2)\]
Correct Answer: A
Solution :
\[{{T}_{n}}={{n}^{2}}+n\]\[\Rightarrow \]\[{{S}_{n}}=\Sigma {{T}_{n}}=\Sigma {{n}^{2}}+\Sigma n\] \[=\frac{n\,(n+1)\,(2n+1)}{6}+\frac{n\,(n+1)}{2}\] \[=\frac{n\,(n+1)}{6}\,\{2n+1+3\}=\frac{n\,(n+1)\,(n+2)}{3}\].
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