Question

find the average of square of the natural number from 1 to 41

Answer 1

sum of squares =n(n+1)(2n+1)/6=41*42*83/6
average=41*42*83/6*41=42*83/6
=7*83

Answer 2

Answer :

We know that sum of the square of natural numbers = n (n+1) (2n+1) / 6

and the Value of n is 41.

So,

Sum of Square is  

=> 41 (41+1) (2*41+1) / 6

=> 41 * 42 (82+1) / 6

=> 41 * 42 * 43 / 6

=> 41 * 7 * 83

=> 287 * 83

=> 23821

Then the average of 41 numbers = sum of numbers/n

=> 23821/ 41 = 581

So, The average of square of the natural number from 1 to 41 is 581.

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