The mean of the following data 12- 22- 32- 42- - n2 isA- -12 n-1B- 6n-12 n-1- C- 6n-162 n-1-6n-12 n-1D -nn-162 n-1-6

The correct option is A The sum of the squares of the first n natural numbers is n(n+1)(2n+1)/6. The mean is the sum divided by the number of terms. Therefore ...

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Standard IX

Mathematics

Mean

The mean of t...

Question

The mean of the following data $1^{2}, 2^{2}, 3^{2}, 4^{2}, . . . . . . . . n^{2}$ is

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Solution

The correct option is A

The sum of the squares of the first n natural numbers is $\frac{n (n + 1) (2 n + 1)}{6}$ .
The mean is the sum divided by the number of terms. Therefore, the mean is

$\frac{\frac{n (n + 1) (2 n + 1)}{6}}{n}$ = $\frac{(n + 1) (2 n + 1)}{6}$

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Similar questions

The mean of the following data $1^{2}, 2^{2}, 3^{2}, 4^{2}, . . . . . . . . n^{2}$ is

Find the sum of $1. n^{2} + 2 (n - 1)^{2} + 3 (n - 2)^{2} + . . . . . . n {.1}^{2}$

$1.3 + 3.5 + 5.7 + . . . . + (2 n - 1) (2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{6}$

$1.3 + 3.5 + 5.7 + \dots + (2 n - 1) (2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3} .$

1.3 + 3.5 + 5.7 + \dots (2 n - 1) (2 n + 1) = \frac{n (4 n + 6 n - 1)}{3}

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The mean of the following data 12,22,32,42,........n2 is

The correct option is A The sum of the squares of the first n natural numbers is n(n+1)(2n+1)6. The mean is the sum divided by the number of terms. Therefore, the mean is n(n+1)(2n+1)6n=(n+1)(2n+1)6

The mean of the following data $1^{2}, 2^{2}, 3^{2}, 4^{2}, . . . . . . . . n^{2}$ is

The correct option is A

The sum of the squares of the first n natural numbers is $\frac{n (n + 1) (2 n + 1)}{6}$ .
The mean is the sum divided by the number of terms. Therefore, the mean is

$\frac{\frac{n (n + 1) (2 n + 1)}{6}}{n}$ = $\frac{(n + 1) (2 n + 1)}{6}$