Statement-1: The variance of first n even natural numbers is n^2-14 Statement-2: The sum o…

Question

Statement-1: The variance of first n even natural numbers is $\dfrac{n^2-1}{4}$
Statement-2: The sum of first n natural numbers is $\dfrac{n(n+1)}{2}$ and the sum of squares of first n natural numbers is $\dfrac{n(n+1)(2n+1)}{6}$.

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.

Solutions · 1