Why Read This Book? - Index of

158 Chapter 4 Functions
By Eq. (4.26), it follows that the entries in row n are C(n, k), where the diagonal
column of 1s headed downward left and downward right correspond to k = 0
and k = n, respectively.
Here’s an important place where the entries in Pascal’s triangle appear. Suppose
you want to expand the expression (a + b) n for some nonnegative integer n.
Rather than multiply out (a + b)(a + b) ···(a + b) using the extended distributive
property, there is a much easier way to see what the terms are, and it involves
C(n, k) in the coefficients.
Theorem 4.10.2 (Binomial Theorem). Let a and b be nonzero real numbers,
and let n be a nonnegative integer. Then
(a + b) n =
=
� �
n
a
0
n b 0 +
n�
k=0
� �
n
a
k
n−k b k =
� �
n
a
1
n−1 b 1 +
n�
k=0
� �
n
a
2
2 b 2 +···+
� �
n
a
n − k
n−k b k
� �
n
a
n
0 b n
(4.32)
The only reason we do not allow a or b to be zero in Theorem 4.10.2 is that
0 0 is not defined, so Eq. (4.32) would produce some undefined terms. If either a
or b is zero, the expansion of (a + b) n is not particularly interesting. Thus we omit
it. Using the entries from Pascal’s triangle, we have
(a + b) 0 = 1a 0 b 0 = 1
(a + b) 1 = 1a 1 b 0 + 1a 0 b 1 = a + b
(a + b) 2 = a 2 + 2ab + b 2
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 , etc.
(4.33)
We can argue the binomial theorem in several ways using combinatorial style
arguments that appeal to some of our previous counting techniques. One way is to
note that applying the extended distributive property to (a + b)(a + b) ···(a + b)
is tantamount to creating a whole bunch of terms of a form like aaabbaab, where
from each (a + b) factor we select either a or b. Since each factor allows for two
possible choices, there are 2 n terms generated, and every one is of the form a n−k b k
for some 0 ≤ k ≤ n. To determine the coefficient of a n−k b k , we must determine how
many times the term a n−k b k appears in all the distribution of the multiplication.
If k of the factors supply us with b and the remaining factors provide us with a,
then the number of times a n−k b k appears in the grand sum is the same as the
number of ways of choosing k of the n terms to provide us with b (or n − k of the
terms to provide us with a). That is, the term a n−k b k is produced C(n, k) times in
the distribution.