A) \[\frac{{{n}^{2}}+1}{12}\]
B) \[\frac{{{n}^{2}}-1}{12}\]
C) \[\frac{(n+1)(2n+1)}{6}\]
D) None of these
Correct Answer: B
Solution :
We have, \[{{\sigma }^{2}}=\frac{1}{n}{{\sum\limits_{i=1}^{n}{x_{i}^{2}-\left( \frac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}} \right)}}^{2}}\] \[=\frac{1}{n}({{1}^{2}}+{{2}^{2}}+...+{{n}^{2}})-{{\left( \frac{1}{n}(1+2+...+n) \right)}^{2}}\] \[=\frac{1}{n}.\frac{n(n+1)(2n+1)}{6}-\frac{1}{{{n}^{2}}}.{{\left[ \frac{n(n+1)}{2} \right]}^{2}}\] \[=\frac{(n+1)(2n+1)}{6}-\frac{{{(n+1)}^{2}}}{4}=\frac{{{n}^{2}}-1}{12}\]
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