Prove the relation: 1² + 2² + 3² ... n² = 1/6 n(n+1)(2n+1)

We can apply the mathematical induction technique to prove
this statement that the sum of the square of the 1st natural numbers is
n(n+1)(2n+1)/6.

The first step:
Basis;

We'll note the identity by
P(n)

We'll prove that the statement holds for
n=1.

P(1): 1^2 =
1*(1+1)*(2+1)/6

1 = 2*3/6 => 1 = 1
true

The next step is the inductive step, where we show
that if P(k) holds, then P(k+1) holds, too.

P(k): 1^2 +2^2
+ ... + k^2 = k*(k+1)(2k+1)/6 true

P(k+1): 1^2 +2^2 + ... +
k^2 + (k+1)^2 = (k+1)[(k+1)+1][2(k+1)+1]/6

But 1^2 +2^2 +
... + k^2 = k*(k+1)(2k+1)/6

P(k+1): k*(k+1)(2k+1)/6 +
(k+1)^2 = (k+1)(k+2)(2k+3)/6

P(k+1): (k+1)[k*(2k+1) + 6k +
6] = (k+1)(k+2)(2k+3)

We'll reduce by
(k+1):

P(k+1): (2k^2 + 7k + 6) = (2k^2 + 3k+4k +
6)

P(k+1): (2k^2 + 7k + 6) = (2k^2 + 7k +
6)

Since both sides are equal, P(k+1)
holds.

Since both steps of mathematical
induction have been proved, the statement P(n): 1^2 + .... + n^2 = n*(n+1)*(2n+1)/6
holds for all natural numbers n.