A) \[\frac{n}{6}(n+1)(6{{n}^{2}}+14n+7)\]
B) \[\frac{n}{6}(n+1)(2n+1)(3n+1)\]
C) \[4{{n}^{3}}+4{{n}^{2}}+n\]
D) None of these
Correct Answer: A
Solution :
This is an A.G. series whose \[{{n}^{th}}\] term is equal to \[{{T}_{n}}=n{{(2n+1)}^{2}}=4{{n}^{3}}+4{{n}^{2}}+n\] \[\therefore \] \[{{S}_{n}}=\sum\limits_{1}^{n}{{{T}_{n}}}=\sum\limits_{1}^{n}{(4{{n}^{3}}+4{{n}^{2}}+n)}\] \[=4\sum\limits_{1}^{n}{{{n}^{3}}}+4\sum\limits_{1}^{n}{{{n}^{2}}}+\sum\limits_{1}^{n}{n}\] \[=4{{\left\{ \frac{n}{2}(n+1) \right\}}^{2}}+\frac{4}{6}n(n+1)(2n+1)+\frac{n}{2}(n+1)\] \[=\frac{n}{6}(n+1)(6{{n}^{2}}+14n+7)\].
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