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