110 
PYTHAGOREAN ARITHMETIC 
Then the gnomon 
BG' D = BB'. BG+DD'. G'D' 
= BB' (BG + G'D'). 
Now BG = \n (n +1), 
G'B' = 1 + 2 + 3 + ... + {n~ 1) = ±n{n— 1), BB' = n\ 
therefore (gnomon BG'D) = n ,n~ = n 2 . 
Similarly (gnomon B'C"D') = {n — l) 3 , and so on. 
Therefore l 3 + 2 3 + ... + n z = the sum of the gnomons round 
the small square at A which has 1 for its side plus that small 
square; that is, 
l 3 + 2 3 + 3 3 + ... + n z — square AC = n (n+ l)} 2 . 
It is easy to see that the first gnomon about the small 
square at A is 3 + 5 = 2 3 , the next gnomon is 7 + 9 + 11 = 3 3 , 
and so on. 
The demonstration therefore hangs together with the 
theorem stated by Nicomachus. Two alternatives are possible. 
Alkarkhi may have devised the proof himself in the Greek 
manner, following the hint supplied by Nicomachus’s theorem. 
Or he may have found the whole proof set out in some 
Greek treatise now lost and reproduced it. Whichever alter 
native is the true one, we can hardly doubt the Greek origin 
of the summation of the series of cubes. 
Nicomachus passes to the theory of arithmetical proportion 
and the various means (cc. 21-9), a description of which has 
already been given (p. 87 above). There are a few more 
propositions to be mentioned under this head. If a-^h = h—c, 
so that a, h, c are in arithmetical progression, then (c. 23. 6) 
h- —ac = (a—h) 2 = (h—c) 2 ,