as
the coefficient of an-rbr in the expansion
of (a + b)n, and tabulated these coefficients
in the arrangement of the Pascal Triangle:
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In Chapter 1 we definedas
the coefficient of an-rbr in the expansion
of (a + b)n, and tabulated these coefficients
in the arrangement of the Pascal Triangle:
| n | Coefficients of (a + b)n | ||||||||||||||
| 0 | 1 | ||||||||||||||
| 1 | 1 | 1 | |||||||||||||
| 2 | 1 | 2 | 1 | ||||||||||||
| 3 | 1 | 3 | 3 | 1 | |||||||||||
| 4 | 1 | 4 | 6 | 4 | 1 | ||||||||||
| 5 | 1 | 5 | 10 | 10 | 5 | 1 | |||||||||
| 6 | 1 | 6 | 15 | 20 | 15 | 6 | 1 | ||||||||
| ... | . | . | . | . | . | . | . | . | |||||||
for n = 0, 1, 2, ... . We also noted that each number inside the
border of 1's is the sum of the two closest numbers on the previous line.
This property may be expressed in the form
This formula provides an efficient method of generating successive lines
of the Pascal Triangle, but the method is not the best one if we want only
the value of a single binomial coefficient for a large n, such as 
We therefore seek a more direct approach.
It is clear that the binomial coefficients in a diagonal adjacent to
a diagonal of 1's are the numbers 1, 2, 3, ... ; that is,
Now let us consider the ratios of binomial coefficients to the previous
ones on the same row. For n = 4, these ratios are:
(2) 4/1, 6/4 = 3/2, 4/6 = 2/3, 1/4.For n = 5, they are
(3) 5/1, 10/5 = 2, 10/10 = 1, 5/10 = 1/2, 1/5.The ratios in (3) have the same pattern as those in (2) if they are rewritten as
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Monday, June 22, 1998