The S.D. of n natural numbers is

$$\begin{gathered} \sum \mathrm{x}=\frac{\mathrm{n}(\mathrm{n}+1)}{2} \sum {\mathrm{x}}^2=\frac{\mathrm{n}(\mathrm{n}+1) (2\mathrm{n}+1)}{6} \\ \mathrm{S}.\mathrm{D}=\sqrt{\frac{\sum {\mathrm{x}}^2}{\mathrm{n}}-{\left(\frac{\sum \mathrm{x}}{\mathrm{n}}\right)}^2} \\ =\sqrt{\frac{\mathrm{n}(\mathrm{n}+1) (2\mathrm{n}+1)}{6\mathrm{n}}-{\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2\mathrm{n}}\right)}^2} \\ =\sqrt{\frac{(\mathrm{n}+1) (2\mathrm{n}+1)}{6}-\frac{{(\mathrm{n}+1)}^2}{4}} \\ =\sqrt{\frac{\mathrm{n}+1}{2} \left(\frac{4\mathrm{n}+2-3\mathrm{n}-3}{6}\right)} \\ =\sqrt{\frac{\mathrm{n}+1}{12} ·(\mathrm{n}-1)} \\ =\sqrt{\frac{{\mathrm{n}}^2-1}{12}} \end{gathered}$$