A) \[\frac{{{n}^{2}}-2n}{3}\]
B) \[\frac{2{{n}^{2}}+n}{3}\]a
C) \[\frac{n(n+2)}{3}\]
D) \[\frac{2{{n}^{2}}-n}{3}\]
Correct Answer: C
Solution :
Given series, \[\frac{{{1}^{2}}}{1}+\frac{{{1}^{2}}+{{2}^{2}}}{1+2}+\frac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}{1+2+3}+....\] The nth terms of the series is \[{{T}_{n}}=\frac{\sum{{{n}^{2}}}}{\sum{n}}=\frac{n(n+1)(2n+1)}{\frac{6n(n+1)}{2}}=\frac{(2n+1)}{3}\] Now, \[{{S}_{n}}=\frac{1}{3}(2\sum{n}+\sum{1})\] \[=\frac{1}{3}\left\{ 2.\frac{n(n+1)}{2}+n \right\}=\frac{n}{3}(n+1+1)\] \[=\frac{n(n+2)}{3}\]
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