A) \[\frac{{{n}^{2}}+1}{12}\]
B) \[\frac{{{n}^{2}}-1}{12}\]
C) \[\frac{(n+1)(2n+1)}{6}\]
D) \[{{\left[ \frac{n(n+1)}{2} \right]}^{2}}\]
Correct Answer: B
Solution :
\[\bar{x}=\frac{1+2+3+...+n}{n}\] \[=\frac{(n+1)}{2}\] \[\therefore \] \[{{\sigma }^{2}}=\frac{\sum {{({{x}_{i}})}^{2}}}{n}-{{(\bar{x})}^{2}}\] \[=\frac{\sum {{n}^{2}}}{n}-{{\left( \frac{n+1}{2} \right)}^{2}}\] \[=\frac{n(n+1)(2n+1)}{6n}-{{\left( \frac{n+1}{2} \right)}^{2}}\] \[=\frac{{{n}^{2}}-1}{12}\]
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