Question

(8 m (a) The sum of squares of the first n natural numbers is given by n(n + 1)(2n + 1) 6 Find the sum of squares of the first 12 natural number​

Answer 1

Given:

The sum of squares of the first n natural numbers is given by  $\frac{n(n+1)(2n+1)}{6}$.

To Find:

Find the sum of squares of the first 12 natural number​?

Step-by-step explanation:

• We have sum of square of n natural numbers is given by the formula

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

• To find sum of squares of the first 12 natural number​ we will put n=12 in above equation.

                         $\frac{12(12+1)(2(12)+1)}{6}$

• Now we will solve this equation by opening bracets we get

                      [tex]\Rightarrow\frac{12(13)(24+1)}{6} \\\\ \Rightarrow \frac{12(13)(25)}{6} \\\\ \Rightarrow\frac{3900}{6} =650[/tex]

Thus sum of square of first 12 terms is 650.