Question

How much will be the average of the squares of natural numbers from 1 to 40?

Answer 1

Step-by-step explanation:

=> sum = 1²+2²+3²+4²+....+40²

formula for square A. P = n(n+1)(2n+1)/6

=> sum = 40*(41)*(81)/3*2

=> sum = 20*41*27

=> sum = 22,140

total terms = 40

=> Average = 22,140/40

=> Avg. = 553.5

Answer 2

Concept:

We need to recall the following formulas to solve this question.

• Sum of squares of first n natural numbers ($1^{2}+2^{2}+3^{2}+........+n^{2}$) is given by : $S_{n} = \frac{n(n+1)(2n+1)}{6}$  .
• Average of numbers = $\frac{sum \;of \;all \;numbers}{total \;number}$ .

Given:

We have given natural numbers from 1 to 40.

To find:

We have to find the average of the squares of given natural numbers.

Solution:

Squares of natural numbers from 1 to 40 are:

$1^{2},2^{2},3^{2},........,40^{2}$.

Total terms = 40

Sum of square of natural numbers is given by:

$S_{40} = \frac{40(40+1)(2\times40+1)}{6}$

      =  $\frac{40\times41(80+1)}{6}$

      =  $\frac{40\times41\times 81}{6}$

      = 20 × 41 × 27

      = 22140

Average =  $\frac{S_{40}}{40}$

              =  $\frac{22140}{40}$

              = 553.5

Hence, The average of squares of natural numbers from 1 to 40 is : 553.5