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).**

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