Prove that
for all natural n.
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Let's start from what we know.



Note that:



(sign of last term will be + when n is even and - when n is odd).
Sum is finite so we can split it into two sums, first with only positive trems (squares of even numbers) and second with negative (squares of odd numbers). So:



And now the proof.

1) n is even.

In this case, both and have terms. For example if n=8 then:



Generally, there will be:



Now, calculate our sum:







So in this case we prove, that:

 

2) n is odd.

Here, has more terms than . For example if n=7 then:



So there is terms in , terms in and:



Now, we can calculate our sum:









We consider all possible n so we prove that: