Question

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:

Put n=10 in the given expression

Sum of squares of first n natural numbers is given by

6

1

n(n+1)(2n+1)

Therefore, the sum of squares of the first 10 natural numbers

=

6

1

10(10+1)(2×10+1)

=

6

1

×10×11×21

=385

Hence, the sum of squares of the first 10 natural numbers is 385.

Answer 2

The sum of squares of first n natural numbers is given by

n(n+1)(2n+1)/6

sum of squares of the first 10 natural numbers.​ = 385

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