Prove that
mean of squares of first n natural numbers is (n+1)(2n+1)/6
you will recall that
n
∑ k^2 = n(n+1)(2n+1)/6
k=1
it's pretty obvious then what the average is
mean of squares of first n natural numbers is (n+1)(2n+1)/6
n
∑ k^2 = n(n+1)(2n+1)/6
k=1
it's pretty obvious then what the average is