College Algebra, 2013a

9.4 The Binomial Theorem 689
( 0
0)
( ) ( 1 1
0
1)
( ) ( 2 2
0
1
↘↙
) ( 2
2)
↘↙
↘↙
) ( 3
3)
( 3
0
) ( ) ( 3 3
1
2
↘↙ ↘↙ ↘↙
) ( ) ( ) ( ) ( 4 4 4 4
1
2
3
4)
( 4
0
.
Since ( (
n
0)
= 1 and
n
n)
= 1 for all whole numbers n, we get that each row of Pascal’s Triangle
begins and ends with 1. To generate the numbers in the middle of the rows (from the third row
(
onwards), we take advantage of the additive relationship expressed in Theorem 9.3. For instance,
1
) (
0 + 1
) (
1 = 2
) (
1 , 2
) (
0 + 2
) (
1 = 3
)
1 and so forth. This relationship is indicated by the arrows in the
array above. With these two facts in hand, we can quickly generate Pascal’s Triangle. We start
with the first two rows, 1 and 1 1. From that point on, each successive row begins and ends with
1 and the middle numbers are generated using Theorem 9.3. Below we attempt to demonstrate
this building process to generate the first five rows of Pascal’s Triangle.
1
1 1
↘↙
1 1+1 1
1
1 1
1 2 1
↘↙ ↘↙
1 1+2 2+1 1
1
1 1
1 2 1
1 3 3 1
↘↙ ↘↙ ↘↙
1 1+3 3+3 3+1 1
−−−−−−→
−−−−−−→
−−−−−−→
1
1 1
1 2 1
1
1 1
1 2 1
1 3 3 1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1