Question

What is the average of square of the natural nbers from 1 to 35?

Answer 1

Answer:

So Average would be 426

Step-by-step explanation:

We are given Natural Numbers from 1 to 35

We have to find their average

Average = Sum / Total no of Numbers

Now Firstly we have to find the Sum

Sum = (1)²+(2)²+(3)²+(4)²+.........................+(35)²

It is sum of the squares of the Numbers

Now by Faulhaber's formula Which we read that Sum of squares of first n natural Numbers are given by the formula

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

Here in our question value of n is 35

so putting it in formula

Sum $=\frac{35*(35+1)*(2(35)+1)}{6}$

Sum $=\frac{35*(36)*(71)}{6}$

Sum $=\frac{89460}{6}$

Sum = 14910

Now

Average $=\frac{Sum}{Total Elements}$

Putting the value

Average $=\frac{14910}{35}$

Average = 426

So Average would be 426

Answer 2

Answer: 426

Step-by-step explanation:

To Find: Average of squares of the natural number 1 to 35

Now as we know

$1^1+2^2+3^2+4^2+........... = \dfrac{n(n+1)(2n+1)}{6}$

where n is number of terms

$Average=\dfrac{\text { sum of observations }}{\text {number of observations}} \\\\=\dfrac{\dfrac{n(n+1)(2n+1)}{6} }{n} \\\\= \dfrac{(n+1)(2n+1)}{6} =\dfrac{ (35+1)(2\times 35+1)}{6} = \dfrac{36\times 71}{6} = 6\times 71 = 426$

Hence, the sum of squares of natural numbers from 1 to 35 is 426