A) \[\frac{n(n+1)(2n+1)}{3}\]
B) \[\frac{2n(n+1)(2n+1)}{3}\]
C) \[\frac{n(n+1)(2n+1)}{6}\]
D) \[\frac{n(n+1)(2n+1)}{9}\]
Correct Answer: B
Solution :
\[{{2}^{2}}+{{4}^{2}}+{{6}^{2}}+........+{{(2n)}^{2}}\] \[={{2}^{2}}({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.......+{{n}^{2}})\] \[=\frac{4n(n+1)(2n+1)}{6}=\frac{2n(n+1)(2n+1)}{3}\].
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