પ્રથમ $ n$  એકી પ્રાકૃતિક સંખ્યાઓના $S.D. $ માટે

$1, 3, 5, ……. , (2n – 1)$

$\sum x_i = 1 + 3 + 5 + …. + (2n – 1) = n^2$

$\sum x_i2 = 12 + 32 + 52 + …… + (2n – 1)^2$

$= [1^2 + 2^2 + 3^2 + ….. + (2n)^2] – [2^2 + 4^2 + …… + (2n)^2]$

$= [1 + 2^2 + 3^2 + ….. + (2n)^2] – 2^2[1^2 + 2^2 + ….. + n^2]$

$ = \,\,\,\frac{{2n(2n\,\, + \,\,1)\,(4n\,\, + \,\,1)}}{6}\,\, – \,\,4\,\,.\,\,\,\frac{{n(n\,\, + \,\,1)\,(2n\,\, + \,\,1)}}{6}$

$ = \,\,\,\frac{{n(2n\,\, + \,\,1)}}{3}\,[(4n\,\, + \,\,1)\,\, – \,\,2(n\,\, + \,\,1)]\,\,\,\,\,\,\,$

$ = \,\,\frac{{n(2n\,\, + \,\,1)\,(2n\,\, – \,\,1)}}{3}\,\, = \,\,\frac{{n(4{n^2}\, – \,\,1)}}{3}$

માંગેલું $S.D. $ $ = \,\,\,\sqrt {\frac{{\Sigma {\text{x}}_{\text{i}}^{\text{2}}}}{{\text{n}}}\,\, – \,\,{{\left( {\frac{{\Sigma {x_i}}}{n}} \right)}^2}} \,\,\,$

$\,\, = \,\,\sqrt {\frac{{4{n^2}\, – \,\,1}}{3}\,\, – \,\,{n^2}} \,\,\,\,\, = \,\,\sqrt {\frac{{{n^2}\, – \,\,1}}{3}} $