SUM OF SERIES OF CUBE NUMBERS 
109 
3 and 
same 
Nicomachus does not give this formula, but it was known 
to the Roman agrimensores, and it would be strange if 
)f the 
form 
tween 
given 
), and 
e, for 
Nicomachus was not aware of it. It may have been dis 
covered by the same mathematician who found out the 
proposition actually stated by Nicomachus, which probably 
belongs to a much earlier time. For the Greeks were from 
the time of the early Pythagoreans accustomed to summing 
the series of odd numbers by placing 3, 5, 7, &c., successively 
as gnomons round 1 ; they knew that the result, whatever 
the number of gnomons, was always a square, and that, if the 
1)> 
number of gnomons added to 1 is (say) r, the sum (including 
the 1) is (r+1) 2 . Hence, when it was once discovered that 
the first cube after 1, i.e. 2 3 , is 3 + 5, the second, or 3 3 , is 
7 + 9 + 11, the third, or 4 3 , is 13 + 15 + 17 + 19, and so on, they 
were in a position to sum the series 1 3 + 2 3 + 3 3 +...+r 3 ; 
for it was only necessary to find out how many terms of the 
series 1 + 3 + 5 + ... this sum of cubes includes. The number 
3. If, 
of terms being clearly 1 + 2 + 3 + ... + r, the number of 
gnomons (including the 1 itself) is \t (r + 1) ; hence the sum 
of them all (including the l), which is equal to 
l 3 + 2 3 + 3 3 + ... + r 3 , 
) is a 
so on. 
to the 
nding 
refore 
is {^r(r+ l)} 2 . Fortunately we possess apiece of evidence 
which makes it highly probable that the Greeks actually 
dealt with the problem in this way. Alkarkhi, the Arabian 
algebraist of the tenth-eleventh century, wrote an algebra 
under the title Al-Fakhrl. It would seem that there were at 
the time two schools in Arabia which were opposed to one 
another in that one favoured Greek, and the other Indian, 
methods. AlkarkhI was one of those who followed Greek 
□nula 
models almost exclusively, and he has a proof of the theorem 
now in question by means of a figure with gnomons drawn 
in it, furnishing an excellent example of the geometrical 
algebra which is so distinctively Greek, 
: + r— 1). 
learly 
-1) 
Let AB be the side of a square AC; let 
AB = 1 + 2 + ... + n = {n + 1), 
and suppose BB' = n, B' B" — n— 1, B" B'" = n— 2, and so on. 
Draw thé squares on AB', AB"... forming the gnomons 
shown in the figure. 
-1)