Question

find. the varience. of first n natural number

Answer 1

Answer:

Step-by-step explanation:

First n natural numbers are 1,2,3,…,n.

Mean=(1+2+3+…+n)÷n

=[n(n+1)/2]/n

(n+1)/2 …(i)

And, 1²+2²+3²+…+n²=n(n+1)(2n+1)/6

Or, (1²+2²+3²+…+n²)/n= {(n+1)(2n+1)/6} …(ii)

Variance of first n natural numbers={(1²+2²+3²+…+n²)/n}-(Mean)²

={(n+1)(2n+1)/6}-{(n+1)²/4}

=(n+1)/12{4n+2–3(n+1)}

=(n+1)(n-1)/12

=(n²-1)/12