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number of kernels was determined as the second power (the square)
of the number of the board square considered. In this case, the pay
ments would proceed as follows:
Square 1 requires 1
2
1 kernel.
Square 2 requires 2
2
4 kernels.
Square 3 requires 3
2
9 kernels.
Square 4 requires 4
2
16 kernels.
Square
i
requires
i
2
kernels.
Square 64 requires 64
2
4096 kernels.
Here, the payments are increasing considerably faster than when
they increase by only two for each square, but the total (1 4 9
16 . . . 4096 89,440 kernels) (about 1.2 bushels) is still
manageable.
Although these alternate formulae yield significantly smaller
payments of corn for Paul, other formulae can give much bigger re
sults. For example,
1.
Start with 1 kernel for square 1.
For square 2, multiply the amount from square 1 by 2.
For square 3, multiply the amount from square 2 by 3.
For square 4, multiply the amount from square 3 by 4.
For square
i
, multiply the amount from square
i
1 by 1.
Note:
Here, one can show that the payment for square
i
is
i
(
i
1) (
i
2) . . . 2 1. This number is sometimes called
i
factorial, and is written
i
! .
2.
For square 1, pay 1
1
kernels.
For square 2, pay 2
2
kernels.
For square 3, pay 3
3
kernels.
For square
i
, pay
i
i
kernels.
Each of these formulae produces much larger payments than were
required of the king in Option 2. (If you're feeling ambitious, try de
termining the size of the payments due in each case for the first row of
squares on a chessboard.)
