$\begin {array} {1 1} (A)\;\large\frac{n^2+1}{12} & \quad (B)\;\large\frac{n^2-1}{12} \\ (C)\;\large\frac{(n+1)(2n+1)}{6} & \quad (D)\;\bigg[ \large\frac{n(n+1)}{2} \bigg]^2 \end {array}$

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$ \overline x = \large\frac{1+2+3+....+n}{n}$

$ = \large\frac{(n+1)}{2}$

$ \therefore \sigma^2 = \large\frac{ \Sigma x_i^2}{n} $ $ - ( \overline x)^2$

$ = \large\frac{ \Sigma n^2}{n}$ $ - \large\frac{(n+1)^2}{2}$

$ \large\frac{(n+1)(2n+1)}{6n} $ $ - \bigg( \large\frac{n+1}{2} \bigg)^2$

$ = \large\frac{n^2-1}{12}$

Ans : (B)

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