The History of Mathematics: A Brief Course

3. COMBINATORICS 211

43,046,721 colors. It does not ask for the sum of this series, which indeed is an

absurd operation, given that the objects are of different kinds.

Hindu mathematicians gave rules for summing geometric progressions and also

the terms of arithmetic progressions and their squares. In Section 3 of Chapter 12 of

the Brahmasphutasiddhanta (Colebrooke, 1817, pp. 290-295), Brahmagupta gives

four rules for dealing with arithmetic progressions. The first rule gives the sum

of an arithmetic progression as its average value (half the sum of its first and last

terms) times its period, which is the number of terms in the progression. We would

write this rule as the formula

Ó

é , , .A / , ,,a + {a + nd) .n(n+ 1)

(a + kd) = (n + 1) ~ = (n + l)a + d-±——-.

fc=o 2 2

For the case á = 0 and d = 1, this formula gives the familiar rule

fc=i^2

and Brahmagupta then says that the sum of the squares will be this number mul-

tiplied by twice the period added to 1 and divided by 3; in other words

ç

Ó*

fc=l

2 n(n +l)(2n+1)

For visual proofs of these results Brahmagupta recommended using piles of balls or

cubes.

These same rules were given in Chapter 5 of Bhaskara's Lilavati (Colebrooke,

1817, pp. 51-57). Bhaskara goes a step further, saying, "The sum of the cubes of

the numbers one, and so forth, is pronounced by the ancients equal to the square

of the addition." This also is the correct rule that we write as

k=l \fc=l / v
Bhaskara also gives the general rule for the sum of a geometric progression of ç + 1
terms (a,ar,ar^2 ,... ,arn) in terms that amount to a(r"+1 —l)/(r—1). He illustrates
this rule with several examples, finding that 2 + 6 + 18 + 54 + 162 + 486 + 1456 =
2(3^7 - l)/2 = 3^7 - 1 = 2186.
About a century later than Bhaskara, Fibonacci's Liber quadratorum gives the
same rule for the sum of the squares (Proposition 10): "If beginning with the unity,
a number of consecutive numbers, both even and odd numbers, are taken in order,
then the triple product of the last number and the number following it and the sum
of the two is equal to six times the sum of the squares of all the numbers, namely
from the unity to the last."
Proposition 11 gives a more elaborate summation rule, which we can express
simply as

(2n + l)(2n + 3)(4n + 4) = 12(l^2 + 3^2 + 5^2 + · • • + (2n + l)^2 )