Question

Variance of 1st n positive integers is

a)(n²-1)/12

b)(n+1)/6

c)(n-1)(2n+1)/3

d)n(n-1)

Answer 1

Sum of first n positive integers = n (n+1)/2

Mean of first n positive integers, μ = (n+1)/2

Variance = ∑ (x_i - μ)²/n

= ((1-μ)²+(2-μ)²+...+(n-μ)²)/n

= (1² + 2²+ ....+n²)/n + μ² - (2μ + 4μ + 6μ +.....+2nμ)/n 

= (n+1)(2n+1)/6 + (n+1)²/4 - μ(n+1)

= (n+1)(2n+1)/6 + (n+1)²/4 - (n+1)²/2

= (n+1)(2n+1)/6 - (n+1)²/4

= (2(n+1)(2n+1) - 3(n+1)²)/12

= (n+1) [4n+2 - 3n -3]/12

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

= (n²-1) / 12