How to prove that (from i=0 to n)Σi^2= (n/6)(n+1)(2n+1), by induction.

First you must show that the statement on the right hand side is true for n=1:
Σ_i=0 i² when n=1, is equal to 1²=1
(1/6)(1+1)(1+2)=(1/6)(2)(3)=1
This means that the statement is true for n=1.
Next you assume that it is true for 'k', where k is any number, and so you get;
Σ_i=0 i² when n=k, is equal to (k/6)(k+1)(2k+2)
You then have to show that the statement is true for n=k+1 which would make;
Σ_i=0 i² when n=k+1, is equal to (k+1)/6(k+2)(2k+3) call this 1)
As the left hand side is a sum, it can be written as;
Σ_i=0 i² when n=k + (k+1)²
We already know the sum of i² when n=k and so we can substitute it in;
(k/6)(k+1)(2k+1) + (k+1)²
We then try and reach 1)
We can factorise out (k+1)
(k+1)[(k/6)(2k+1) +k+1]
Next, multiply the inner brackets;
(k+1)[2k²/6+k/6 +k+1]
Take out a factor of 1/6
(k+1)/6(2k²+k+6k+6)= (k+1)/6(2k²+7k+6)
Finally, factorise the inner bracket;
(k+1)/6(k+2)(2k+3)
As this is equal to 1), we have proven that the statement is true for all values of n.