Sum of the squares of the first n natural numbers - Formula derivation

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Hello, Everyone. In this, we are going to derive the formula to find the sum of the squares of the first n natural numbers.
Let S = 12 + 22 + 32 + .......... + (n-1)2 + n2
(m+1)3 - m3 = (m3 + 3m2 + 3m + 1) - m3 = 3m2 + 3m + 1 ------------ (i)
put m = 1,2,3, ....... , (n-1), n in (i)
Then,
23 - 13  = 3(1)2 + 3(1) + 1 ------------ (1)
33 - 23  = 3(2)2 + 3(2) + 1 ------------ (2)
43 - 33  = 3(3)2 + 3(3) + 1 ------------ (3)
...................
n3 - (n-1)3  = 3(n-1)2 + 3(n-1) + 1 ------------ (n-1)
(n+1)3 - n3  = 3(n)2 + 3(n) + 1 ------------ (n)

Now, by adding equations (1),(2),(3), ...... ,(n-1) and (n), we get the following equation.
(n+1)3 - 13  = 3(12 + 22 + 32 + .......... + (n-1)2 + n2) + 3(1 + 2 + 3 + ......... + (n-1) + n) + (1 + 1 + 1 +........ n times)

1 + 2 + 3 + .......... + (n-1) + n = (n(n+1))/2 [Since sum of the first n natural numbers = (n(n+1))/2] and (n+1)3 - 13 = (n3 + 3n2 + 3n + 1) - 13 = n3 + 3n2 + 3n 

Therefore, (n+1)3 - 13 = n3 + 3n2 + 3n = 3S + 3((n(n+1))/2) + n
2n3 + 6n2 + 6n =  6S + 3n2 + 3n + 2n = 6S + 3n2 + 5n
2n3 + 3n2 + n = 6S
n(2n2 + 3n + 1) = 6S
n(2n2 + 2n + n + 1) = 6S
n(2n(n + 1) + 1(n + 1)) = 6S
n(n + 1)(2n + 1) = 6S
Therefore,
S = Sum of the squares of the first n natural numbers = [n(n + 1)(2n + 1)] / 6

Therefore, formula to find the sum of the squares of the first n natural numbers is,

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