The variance of first n natural number is:

Answer

variance =`frac{n^{2}-1}{12}`

Explanation

Variance `=\sum_{x^{2} / n}-(\sum_{x / n})^{2}`


$$ But \sum x=1+2+3+\ldots+n\\ =n\left(n+\frac{1}{2}\right)\\ \sum x^{2}=n(n+1) \frac{(2 n+1)}{6}\\ $$
Substituting these values we get
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