The relation to be proved is : 1^2 + 2^2 + 3^2 ... n^2 = 1/6 n(n+1)(2n+1).

This can be done using mathematical induction.

For n = 1

S1 = 1^2 = 1 and 1*(1 + 1)(2*1 + 1)/6 = 2*3/6 = 1

Let the relation be true for n.

So we have Sn = 1^2 + 2^2 + 3^2 ... n^2 = 1/6 n(n+1)(2n+1)

Now S(n + 1) = (1/6)n(n+1)(2n+1) + (n + 1)^2

=> (n + 1)[n*(2n + 1)/6 + n + 1]

=> (1/6)(n + 1)[ 2n^2 + n + 6n + 6]

=> (1/6)(n + 1)[2n^2 + 7n + 6]

=> (1/6)(n + 1)[2n^2 + 4n + 3n + 6]

=> (1/6)(n + 1)[2n(n + 2) + 3(n + 2)]

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

We now have the relation true for n = 1 and if the relation is true for any n greater than 1, it it also true for n + 1.

**This proves that for all values of n 1^2 + 2^2 + 3^2 ... n^2 = 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.**