JEE MAIN - Mathematics (2009 - No. 1)
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 $${{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}$$
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 $${{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}$$
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
Explanation
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.
$$\therefore$$ 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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