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