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

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First you must show that the statement on the right hand side is true for n=1:

Σi=0 iwhen 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;

Σi=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;

Σi=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;

Σi=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.

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