A) \[\sqrt{\frac{n(n+1)}{2}}\]
B) \[\sqrt{\frac{n(n+1)(2n+1)}{3}}\]
C) \[\sqrt{\frac{n(n+1)}{3}}\]
D) \[\sqrt{\frac{n(n-1)}{2}}\]
E) \[2n+1\]
Correct Answer: C
Solution :
The given series is \[1,\text{ }2,\text{ }3,\text{ }...,\text{ }(2n+1)\] \[\overline{x}=\frac{1+2+3+....+(2n+1)}{2n+1}\] \[=(n+1)\] \[\therefore \]\[{{\sigma }^{2}}=\frac{1}{2n+1}\sum\limits_{r=0}^{2n}{{{\{(1+r)-(1+n)\}}^{2}}}\] \[\left[ \because {{\sigma }^{2}}=\frac{1}{n}\sum{{{({{x}_{i}}-\overline{x})}^{2}}} \right]\] \[=\frac{2}{2n+1}({{1}^{2}}+{{2}^{2}}+....+{{n}^{2}})\] \[\Rightarrow \] \[{{\sigma }^{2}}=\frac{n(n+1)}{3}\] \[\Rightarrow \] \[\sigma =\sqrt{\frac{n(n+1)}{3}}\]
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