# Find the variance of first n even natural numbers ?

## Solution :

$$\therefore$$ Variance = [$${1\over n}\sum{(x_i)^2}$$] – $$(\bar{x})^2$$

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

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

= $$4\over n$$ $$n(n + 1)(2n + 1)\over 6$$ – $$(n+1)^2$$

=  $$(n + 1)[(2n + 1) – 3(n + 1)\over 3$$

= $$(n + 1)(n – 1)\over 3$$ = $$n^2 – 1\over 3$$

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