Question

derive the sum of the squares of the first n natural numbers​

Answer 1

Let us assume the required sum = S

Therefore, S = 12 + 22 + 32 + 42 + 52 + ................... + n2

Now, we will use the below identity to find the value of S:

n3 - (n - 1)3 = 3n2 - 3n + 1

Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get

13 - 03 = 3 . 12 - 3 ∙ 1 + 1

23 - 13 = 3 . 22 - 3 ∙ 2 + 1

33 - 23 = 3 . 32 - 3 ∙ 3 + 1

43 - 33 = 3 . 42 - 3 ∙ 4 + 1

......................................

n3 - (n - 1)3 = 3 ∙ n2 - 3 ∙ n + 1

____ _____

Adding we get, n3 - 03 = 3(12 + 22 + 32 + 42 + ........... + n2) - 3(1 + 2 + 3 + 4 + ........ + n) + (1 + 1 + 1 + 1 + ......... n times)

⇒ n3 = 3S - 3 ∙ n(n+1)2 + n

⇒ 3S = n3 + 32n(n + 1) – n = n(n2 - 1) + 32n(n + 1)

⇒ 3S = n(n + 1)(n - 1 + 32)

⇒ 3S = n(n + 1)(2n−2+32)

⇒ 3S = n(n+1)(2n+1)2

Therefore, S = n(n+1)(2n+1)6

i.e., 12 + 22 + 32 + 42 + 52 + ................... + n2 = n(n+1)(2n+1)6

Thus, the sum of the squares of first n natural numbers = n(n+1)(2n+1)6

Answer 2

Answer:

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