Statement 1: The variance of first n odd natural numbers is n^2-13 Statement 2: The sum of…

Question

Statement 1: The variance of first n odd natural numbers is $\dfrac{n^2-1}{3}$
Statement 2: The sum of first n odd natural numbers is $n^2$ and the sum of squares of first n odd natural numbers is $\dfrac{n(4n^2+1)}{3}$

Accepted Answer

The sum of squares of first n odd numbers is actually $\dfrac{n(4n^2-1)}{3}$, not $\dfrac{n(4n^2+1)}{3}$ as claimed --- so Statement 2 (as stated) is false.

Using the correct formula: $	ext{Var} = \dfrac{4n^2-1}{3}-n^2 = \dfrac{n^2-1}{3}$, confirming Statement 1 is true.

Solutions · 1