Assertion (A): The variance of the first n od | vedclass

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(A) The sum of the first $n$ odd natural numbers is given by $\sum_{i=1}^{n} (2i-1) = n^2$.The sum of the squares of the first $n$ odd natural numbers

Vedclass · Accepted Answer

$(A)$ and $(R)$ are true. $(R)$ is the correct explanation of $(A)$

Vedclass · Answer

(A) The sum of the first $n$ odd natural numbers is given by $\sum_{i=1}^{n} (2i-1) = n^2$.The sum of the squares of the first $n$ odd natural numbers is $\sum_{i=1}^{n} (2i-1)^2 = \sum_{i=1}^{n} (4i^2 - 4i + 1) = 4 \sum i^2 - 4 \sum i + \sum 1$.Using standard summation formulas: $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum i = \frac{n(n+1)}{2}$.Sum $= 4 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 4 \left[ \frac{n(n+1)}{2} \right] + n = \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} = \frac{n(4n^2-1)}{3}$.Thus,Reason $(R)$ is true.Variance is defined as $\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$.$\sigma^2 = \frac{n(4n^2-1)}{3n} - \left( \frac{n^2}{n} \right)^2 = \frac{4n^2-1}{3} - n^2 = \frac{4n^2-1-3n^2}{3} = \frac{n^2-1}{3}$.Thus,Assertion $(A)$ is true and $(R)$ is the correct explanation of $(A)$.

$X$	$5$	$6$	$7$	$8$	$10$
Frequency	$3$	$7$	$4$	$2$	$4$

Class Interval	$0$-$4$	$4$-$8$	$8$-$12$	$12$-$16$	$16$-$20$
Frequency	$2$	$4$	$6$	$3$	$1$

Assertion $(A)$: The variance of the first $n$ odd natural numbers is $\frac{n^2-1}{3}$.
Reason $(R)$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2-1)}{3}$.
Which of the following alternatives is correct?

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Assertion $(A)$: The variance of the first $n$ odd natural numbers is $\frac{n^2-1}{3}$.Reason $(R)$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2-1)}{3}$.Which of the following alternatives is correct?