sum(k^2) = sum(k) + sum(k(k-1)).
To see why this is true, observe that k^2 can be manipulated in the following way:
k^2 = kk = k(k+1-1) = k(1+(k-1)) = k+k(k-1),
sum(k^2) = sum( k+k(k-1)).
Now the question is, can you distribute the sum ( )? Yes of course; observe that
sum(k+k(k-1)) = (1 + 1(1-1)) + (2 + 2(2-1)) + (3 + 3(3-1)) + ...
and from each expression in the parenthesis, we can group all the first terms together and then all the second terms together to get
sum(k+k(k-1)) = (1 + 2 + 3 + ...) + (1(1-1) + 2(2-1) + 3(3-1) + ...)
where we can see that
sum(k+k(k-1)) = sum(k) + sum(k(k-1)).
Then since sum(k^2) = sum(k+k(k-1)), we have
sum (k^2) = sum(k) + sum(k(k-1)).
Note that I merely explained why this is true, I didn't really prove it for the most part. Now another one of Al Kharaji's claims was
sum(k^3) = (sum(k))^2.
This one I will prove using mathematical induction (one of my favorite proof methods). To make the proof easier to read, I typed it up and put it on a pdf file. You can't really put pdf files on blogger.com so I inserted them as pictures. To see them, just hold the "control" key and push the "+" key a few times to zoom in. Then to zoom out, hold the "control" key and push the "-" key, sorry about that.
I probably missed on some minor subtitles, but the main part of the proof should be fine. Now proving this wasn't so bad, it was fun. Make sure to observe the "note" at the end of the proof. I had to actually use the sum of integers formula (lemma 1) to prove sum(k^3) = (sum(k))^2 (lemma 1 actually serves as my daily 5 assignment). I imagine that to prove the sum of squares or sum of cubes formulas, you could use mathematical induction and the proofs would be straightforward; but, let's save those for a different post. I hope we explore more series in the future. If we do, expect more proofs in my blogs. :)