QJEEM09Maaf75: JEE Main 2009 Mathematics - Class 11 JEEM09Maaf75 Solution: Statement - 1 : The variance of first n even natural numbers is …

Question

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Statement - 1 : The variance of first n even natural numbers is

{{{n^2} - 1} \over 4}

Statement - 2 : The sum of first n natural numbers is

{{n\left( {n + 1} \right)} \over 2}

and the sum of squares of first n natural numbers is

{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}

A Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
B Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
C Statement-1 is true, Statement-2 is false
D Statement-1 is false, Statement-2 is true

Accepted Answer

Let first n even natural numbers = 2,4, 6, 8 ...... 2n

\therefore

Sum of those num = 2 + 4 + 6 + ..... 2n

= 2 (1 + 2 + ..... n)

=

2.{{n\left( {n + 1} \right)} \over 2}

= n (n + 1)

\therefore\,\,\,

Mean

\left( {\overline x } \right) = {{n\left( {n + 1} \right)} \over n}

= n + 1

\therefore\,\,\,

Variance

= {1 \over n}\sum {x_i^2} - {\left( {\overline x } \right)^2}

= {1 \over n}\left[ {{2^2} + {4^2} + ...... + {{\left( {2n} \right)}^2}} \right] - {\left( {n + 1} \right)^2}

= {1 \over n}{2^2}\left[ {{1^2} + {2^2} + .......n{}^2} \right] - \left( {n + 1} \right){}^2

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

= {{\left( {n + 1} \right)\left[ {2\left( {2n + 1} \right) - 3\left( {n + 1} \right)} \right]} \over 3}

= {{\left( {n + 1} \right)\left( {n - 1} \right)} \over 3}

= {{{n^2} - 1} \over 3}

\therefore

Statement 1 is false.

Statement 2 is true as those are standard formula.

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