Could someone prove the rule that the sum of n squares is given by : [n(n + 1)(2n + 1)]/6

We have to prove the sum of the squares of the first n
numbers is given by n(n+1)(n+2)/6

We know that (x+1)^3 =
x^3 + 3x^2 + 3x + 1, now if we write the cubes of the numbers 1 to n+1, we
get

1^3 = (0+1) ^3 = 0^3 + 3(0^2) + 3 (0) +
1

2^3 = (1+1) ^3 = 1^3 + 3(1^2) + 3 (1) +
1

3^3 = (2+1) ^3 = 2^3 + 3(2^2) + 3 (2) +
1

…

n^3 = (n-1+1) ^3 = (n-1)
^3 + 3(n-1) ^2 + 3(n-1) + 1

(n+1)^3 = n^3 + 3n^2 + 3n +
1

We see that for all the cubes above other than (n+1)^3
there are equal terms on the right hand side as well as the left hand side. So we cancel
them and we are left with:

(n+1)^3 = 3*sum of squares from
1 to n + 3*(sum of numbers from 1 to n) + n + 1

Denote the
sum of the squares which we are finding by S

=> n^3
+ 3n^2 + 3n + 1 = 3*S + 3*[sum of numbers from 1 to n] +
n+1

As we are trying to find the sum of the squares I
assume the relation for the sum of the numbers from 1 to n is known, which is
n*(n+1)/2

So n^3 + 3n^2 + 3n + 1 = 3*S + 3*n(n+1)/2 +
n+1

=> n^3 + 3n^2 + 3n + 1 = 3*S + 3*(n^2 + n)/2 +
n+1

=> 3*S = n^3 + 3n^2 + 3n + 1 - 3*(n^2 + n)/2 –
n-1

=> 3*S = n^3 + 3n^2 + 3n + 1 – (3*n^2 -3n)/2 – n
– 1

=> 3*S = n^3 + 3n^2/2 +
n/2

=> S = (1/6)( 2n^3 + 3n^2 +
n)

=> S = (1/6)*n*(2n^2 + 3n
+1)

=> S = (1/6)*n*(2n^2 + 2n + n
+1)

=> S =
(1/6)*n*(2n(n+1)+1(n+1))

=> S =
(1/6)*n*(n+1)(2n+1)

Therefore we prove that
the sum of the squares of the numbers from 1 to n is given by
(1/6)*n*(n+1)(2n+1)