College Algebra, 2013a

9.4 The Binomial Theorem 685
(a + b) 4 =
=
=
4∑
j=0
( 4
j)
a 4−j b j
( 4
0)
a 4−0 b 0 +
( ( ( ( 4 4 4 4
a
1)
4−1 b 1 + a
2)
4−2 b 2 + a
3)
4−3 b 3 + a
4)
4−4 b 4
( ( ( ( ( 4 4 4 4 4
a
0)
4 + a
1)
3 b + a
2)
2 b 2 + ab
3)
3 + b
4)
4
We forgo the simplification of the coefficients in order to note the pattern in the expansion. First
note that in each term, the total of the exponents is 4 which matched the exponent of the binomial
(a+b) 4 . The exponent on a beginsat4anddecreasesbyoneaswemovefromonetermtothenext
while the exponent on b starts at 0 and increases by one each time. Also note that the binomial
coefficients themselves have a pattern. The upper number, 4, matches the exponent on the binomial
(a + b) 4 whereas the lower number changes from term to term and matches the exponent of b in
that term. This is no coincidence and corresponds to the kind of counting we discussed earlier. If
we think of obtaining (a + b) 4 by multiplying (a + b)(a + b)(a + b)(a + b), our answer is the sum of
all possible products with exactly four factors - some a, some b. If we wish to count, for instance,
thenumberofwaysweobtain1factorofb out of a total of 4 possible factors, thereby forcing the
remaining 3 factors to be a, the answer is ( (
4
1)
. Hence, the term
4
1)
a 3 b is in the expansion. The
other terms which appear cover the remaining cases. While this discussion gives an indication as
to why the theorem is true, a formal proof requires Mathematical Induction. 4
To prove the Binomial Theorem, we let P (n) be the expansion formula given in the statement of
the theorem and we note that P (1) is true since
(a + b) 1 ? =
1∑
a + b
?
=
j=0
( 1
j)
a 1−j b j
( 1
0)
a 1−0 b 0 +
a + b = a + b ̌
( 1
1)
a 1−1 b 1
NowweassumethatP (k) is true. That is, we assume that we can expand (a + b) k using the
formula given in Theorem 9.4 and attempt to show that P (k +1)istrue.
4 and a fair amount of tenacity and attention to detail.