A) \[\frac{{{n}^{3}}{{(n+1)}^{3}}(2n+1)}{24}\]
B) \[\frac{n(n+1)(3{{n}^{2}}+7n+2)}{12}\]
C) \[\frac{n(n+1)}{6}[n(n+1)+(2n+1)]\]
D) \[\frac{n(n+1)}{12}[6n(n+1)+2(2n+1)]\]
Correct Answer: B
Solution :
\[{{T}_{n}}={{n}^{2}}(n+1)={{n}^{3}}+{{n}^{2}}\] \[{{S}_{n}}=\Sigma {{T}_{n}}=\Sigma {{n}^{3}}+\Sigma {{n}^{2}}\]\[={{\left[ \frac{n\,(n+1)}{2} \right]}^{2}}+\frac{n\,(n+1)\,(2n+1)}{6}\] \[=\frac{n\,(n+1)}{2}\left[ \frac{n\,(n+1)}{2}+\frac{2n+1}{3} \right]\]\[=\frac{n\,(n+1)(3{{n}^{2}}+7n+2)}{12}\].
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