Statement -1: The variance of first n even natural numbers is \;n²-1/4 Statement -2 : The sum of first n natural numbers is \;n(n+1)/2 and the sum of squares of first n natural numbers is \;n(n+1)(2n+1)/6 [AIEEE 2009]

Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1
Statement- 1 is true, Statement- 2 is true; Statement- 2 is not a correct explanation for Statement-1
Statement- 1 is true, Statement-2 is false
Statement- 1 is false, Statement- 2 is true

Solution:

Let first n even natural numbers

=2,4,6,8 \dots 2n

\therefore

Sum of those num

=2+4+6+\dots+2n

=2(1+2+\dots+n)

=2 \cdot \;\frac{n(n+1)}{2}

=n(n+1)

\therefore \;\;

Mean

(\overline{x})=\;\frac{n(n+1)}{n}=n+1

\therefore

Variance

=\;\frac{1}{n} \sum x_i^2-(\overline{x})^2

=\;\frac{1}{n} [2^2+4^2+\dots+(2n)^2] -(n+1)^2

=\;\frac{1}{n} 2^2 [1^2+2^2+\dots+n^2] -(n+1)^2

\;=\;\frac{4}{n} \left[ \;\frac{n(n+1)(2n+1)}{6} \right] -(n+1)^2

\;=\;\frac{(n+1)[2(2n+1)-3(n+1)]}{3}

\;=\;\frac{(n+1)(n-1)}{3}

\;=\;\frac{n^2-1}{3}

\therefore

Statement 1 is false.

Statement 2 is true as those are standard formula.

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