Question

(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 numbers​

Answer 1

Answer:

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Step-by-step explanation:

The sum of squares of the first n natural numbers is given by :-

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

We are given to find the sum of squares of the first 12 natural numbers.

Natural numbers are part of the number system which are the positive integers starting from 1 till infinity. There are infinite natural numbers.

The first 12 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Sum of squares of first ‘n’ natural numbers is :-

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

Here the value of n is 12.

Therefore, their sum of squares is :-

$\frac{12(12+1)(2(12)+1)}{6}\\=> \frac{12\times13\times(24+1)}{6} \\=> \frac{12\times13\times25}{6} \\=>\frac{3900}{6}\\=>650$

Sum of squares of the first 12 natural numbers is 650.

Answer 2

Answer:

n(n+1)(2n+1) =6

Put 1 in place of n:

1 (1+1) (2n+1) = 6

1 (2) (2n+1) = 6

2 (2n+1) = 6

4n+2 = 6

4n = 6-2

4n = 4

n = 4/4

n = 1

So, the sum of squares of the first 12 natural numbers is 1...

Checking that the answer is correct or incorrect..

Put 1 in place of n

1 (1+1) [2(1)+1] = 6

1 (2) (2+1) = 6

2 (3) = 6

6 = 6

so, the answer is correct.