Solution :
\(\therefore\) Variance = [\({1\over n}\sum{(x_i)^2}\)] – \((\bar{x})^2\)
= \(1\over n\)[\(2^2 + 4^2 + ….. + (2n)^2\)] – \((n+1)^2\)
= \(1\over n\)\(2^2 [ 1^2 + 2^2 + ….. + n^2]\) – \((n+1)^2\)
= \(4\over n\) \(n(n + 1)(2n + 1)\over 6\) – \((n+1)^2\)
= \((n + 1)[(2n + 1) – 3(n + 1)\over 3\)
= \((n + 1)(n – 1)\over 3\) = \(n^2 – 1\over 3\)