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The Binomial Theorem
Lecture 29
Section 6.7
Mon, Apr 3, 2006
The Binomial Theorem
Theorem: Given any numbers a and b and any nonnegative integer n,
(a + b)n = k = 0..n C(n, k)an – kbk. Proof: Use induction on n. Basic step: Let n = 0. Then
(a + b)0 = 1. k=0..0 C(0, k)a0 – kbk = C(0, 0)a0b0 = 1.Therefore, the statement is true when n = 0.
Proof, continued
Inductive stepSuppose the statement is true for some n
0.Then
n
k
kknn
k
kkn
n
k
kkn
nn
bak
nba
k
n
bak
nba
bababa
0
1
0
1
0
1
Proof, continued
n
k
nkknn
n
k
n
k
nkknkknn
n
k
n
k
nkknkknn
bbak
n
k
na
bbak
nba
k
na
bbak
nba
k
na
1
111
1 1
1111
1
1
0
1111
1
1
Proof, continued
Therefore, the statement is true for n + 1. Thus, the statement is true for all n 0.
.1
1
1
0
1
1
111
n
k
kkn
n
k
nkknn
bak
n
bbak
na
Example: Binomial Theorem
Expand (a + b)8.C(8, 0) = C(8, 8) = 1.C(8, 1) = C(8, 7) = 8.C(8, 2) = C(8, 6) = 28.C(8, 3) = C(8, 5) = 56.C(8, 4) = 70.
Example: Binomial Theorem
Therefore,
(a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3
+ 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8.
Example: Calculating 1.016
Compute 1.018 on a calculator. What do you see?
Example: Calculating 1.016
1.018 = (1 + 0.01)8
= 1 + 8(0.01) + 28(0.01)2 + 56(0.01)3
+ 70(0.01)4 + 56(0.01)5 + + 28(0.01)6 + 8(0.01)7 + (0.01)8
= 1 + .08 + .0028 + .000056 + .00000070
+ .0000000056 + .000000000028 + + .00000000000008
+ .0000000000000001
= 1.0828567056280801.
Example: Approximating (1+x)n
Theorem: For small values of x,
and so on.
.6
)2)(1(
2
)1(11
.2
)1(11
.11
32
2
xnnn
xnn
nxx
xnn
nxx
nxx
n
n
n
Example
For example,
(1 + x)8 1 + 8x + 28x2
when x is small. Compute the value of (1 + x)8 and the
approximation when x = .03.
Newton’s Generalization of the Binomial Theorem
Theorem: Given any numbers a, b, and n,
(a + b)n = k=0.. C(n, k)an – kbk.
where
C(n, k) = [n(n – 1)…(n – k + 1)]/k! Note that n need not be an integer nor
positive.
Newton’s Generalization
Expand (a + b)-1 in a series, showing the first 5 terms.
Compute the first 5 coefficientsC(-1, 0) = 1.C(-1, 1) = -1.C(-1, 2) = (-1)(-2)/2! = 1.C(-1, 3) = (-1)(-2)(-3)/3! = -1.C(-1, 4) = (-1)(-2)(-3)(-4)/4! = 1.
Newton’s Generalization
Therefore,
...1
...1
...
4321
44332211
453423211
a
b
a
b
a
b
a
ba
babababaa
babababaaba
Newton’s Generalization
Expand (a + b)5/2 in a series, showing the first 5 terms.
.8
15
!223
25
2,2
5
.2
51,
2
5
.10,2
5
C
C
C
.128
5
384
15
!421
21
23
25
4,2
5
.16
5
48
15
!321
23
25
3,2
5
C
C
Newton’s Generalization
Therefore,
...128
5
16
5
8
15
2
51
...128
5
16
5
8
15
2
51
...128
5
16
5
8
15
2
5
4322/5
44332212/5
42/332/122/12/32/52/5
a
b
a
b
a
b
a
ba
babababaa
babababaaba
Newton’s Generalization
If 0 < b < a, then this series will converge rapidly.
Newton’s Generalization
Approximate (1.2)5/2. Let a = 1 and b = 0.2 = 1/5. Then b/a = 1/5.
.5774375.116000
2523916000
1
400
1
40
3
2
11
...5
1
128
5
5
1
16
5
5
1
8
15
5
1
2
512.1
4322/5
The Multinomial Theorem
Theorem: In the expansion of
(a1 + … + ak)n,
the coefficient of a1n1a2
n2…aknk is
n!/(n1!n2!…nk!)
where n = n1 + n2 + … + nk.
Example: The Multinomial Theorem
Expand (a + b + c + d)3. The terms are
a3, b3, c3, d3, with coefficient 3!/3! = 1.a2b, a2c, a2d, ab2, b2c, b2d, ac2, bc2, c2d,
ad2, bd2, cd2, with coefficient 3!/(1!2!) = 3.abc, abd, acd, bcd, with coefficient 3!/(1!1!
1!) = 6.
Example: The Multinomial Theorem
Therefore,
(a + b + c + d)3 = a3 + b3 + c3 + d3 + 3a2b
+ 3a2c + 3a2d + 3ab2 + 3b2c + 3b2d
+ 3ac2 + 3bc2 + 3c2d + 3ad2 + 3bd2
+ 3cd2 + 6abc + 6abd + 6acd + 6bcd. Find (a + b + c)4.
Actuary Exam Problem
If we expand the expression
(a + 2b + 3c)4,
what will be the sum of the coefficients?