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 the above
Correct Answer: A
Solution :
Given series is \[{{1.3}^{2}}+{{2.5}^{2}}+{{3.7}^{2}}+...\text{ }\infty \] This is an arithmetic-geometric series whose nth term is equal to \[{{T}_{n}}=n{{(2n+1)}^{2}}+4{{n}^{3}}+4{{n}^{2}}+n\] \[\therefore \]\[{{S}_{n}}=\sum\limits_{1}^{n}{{{T}_{n}}}=\sum\limits_{1}^{n}{(4{{n}^{3}}+4{{n}^{2}}+n)}\] \[=4\sum\limits_{1}^{n}{{{n}^{3}}}+4\sum\limits_{1}^{n}{{{n}^{2}}}+\sum\limits_{1}^{n}{n}\] \[=4{{\left( \frac{n}{2}(n+1) \right)}^{2}}+\frac{4}{6}n(n+1)(2n+1)+\frac{n}{2}(n+1)\] \[=n(n+1)\left[ {{n}^{2}}+n+\frac{4}{6}(2n+1)+\frac{1}{2} \right]\] \[=\frac{n}{6}(n+1)(6{{n}^{2}}+14n+7)\]
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