Question

how to calculate average of square of first 10 natural numbers

Answer 1

Find their sum and divide it by the no. of he natural numbers.

Let there is asked to find the average of squares of first n natural numbers.

First find their sum.

$1^2 + 2^2 + 3^2 + ...... + n^2 = \frac{n(n + 1)(2n + 1)}{6}$

This is the equation to find the sum of squares of first n natural numbers.

Then divide the sum by n. (Here the no. of natural numbers is n.)

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

∴ The average of the squares of first n natural numbers, is the quotient obtained by dividing the product of the natural number after n and the natural number after twice of n by 6.

Consider n = 10.

$\frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2}{10} \\ \\ = \frac{1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100}{10} \\ \\ = \frac{385}{10} = 38.5$

By using the equation,

$\frac{(10 + 1)(2 \times 10 + 1)}{6} \\ \\ = \frac{11(20 + 1)}{6} \\ \\ = \frac{11 \times 21}{6} = \frac{231}{6} = 38.5$

38.5 is the average.

Hope this may be helpful.

Please mark my answer as the brainliest if this may be helpful.

Thank you. Have a nice day.

Answer 2

There is no question hard in maths you just need to practice dear