Binomial Theorem

1
Binomial Theorem
2
Case Study
2.1 Summation Notation
2.2 Binomial Theorem
Chapter Summary
2
Case Study
Suppose you are asked to calculate the value of
(1.01)n where n is a positive integer, without
using a calculator.
The student thinks it can be solved by
considering the expansion of (1 ? 0.01)n step by
step. Do you think this is possible?
If the power of the expansion (1 ? 0.01)n is
small, we can still find the sum or the values of
individual terms by expanding it term by term.
Actually, there is an alternative method to
tackle such a problem, which is especially
effective when the power is large. This method
will be discussed in this chapter.
3
2.1 Summation Notation
In mathematics, a simple way to denote the sum of
the terms of a sequence is by using summation
notation.
It involves the symbol S which is a Greek
letter read as sigma.
For example, consider a sequence with general
term T(n). The series T(1) T(2) ... T(n)
can be represented by the notation
The notation is read as the summation of T(i) as
i goes from 1 to n.
4
2.1 Summation Notation
Example 2.1T
Solution
(a)
(b)
5
2.1 Summation Notation
Example 2.2T
Solution
Note that

?
?
6
2.2 Binomial Theorem
A. Pascals Triangle
When we expand the expression (x ? y)n for a
positive integer n, it is too troublesome to use
long multiplication, especially when n is very
large.
Consider (x ? y)3 ? (x ? y)2(x ? y) ? (x2 ? 2xy
? y2)(x ? y)
? (x ? y)3 ? x3 ? 3x2y ? 3xy2 ? y3
Using long multiplication again, (x ? y)4 ? (x ?
y)3(x ? y)
? (x3 ? 3x2y ? 3xy2 ? y3)(x ? y)
? (x ? y)4 ? x4 ? 4x3y ? 6x2y2 ? 4xy3 ? y4
7
2.2 Binomial Theorem
A. Pascals Triangle
The expansion (x ? y)n follows some special
pattern.
Binomial Expansion (x ? y)0 (x
? y)1 (x ? y)2 (x ? y)3
(x ? y)4 (x ? y)5
Coefficients of the
expansion 1 1 1 1 2 1 1 3
3 1 1 4 6 4 1 1 5 10 10 5 1
The above triangle is called Pascals Triangle.
Besides 1, every number equals the sum of the two
numbers to the left and right of it in the
preceding row.
For example
8
2.2 Binomial Theorem
A. Pascals Triangle
In general, besides the above relationship
between the coefficients, for n ? 2, the
properties of the expansion of (x ? y)n are
stated as follows
1. There are (n ? 1) terms. 2. When the powers
of x in the expansion decrease by 1, the powers
of y will increase by 1. 3. The sum of the
powers of x and y in each term must be equal to
n. 4. The coefficients of the first term and the
last term are both 1. 5. The coefficients show
a symmetrical pattern.
9
2.2 Binomial Theorem
A. Pascals Triangle
Example 2.3T
Using the properties of Pascals Triangle, expand
the following expressions. (a) (2 ? ax)4 (b)
(x2 ? 1)5
Solution
(a) (2 ? ax)4
(b) (x2 ? 1)5
10
2.2 Binomial Theorem
B. Binomial Theorem
Pascals Triangle is very useful for expanding (x
? y)n when n is small.
However, in the case when n is large, such as 30
or more, the work will be tedious.
Recall the definition of the notation C n
r
Rewrite Pascals Triangle using the notation
11
2.2 Binomial Theorem
B. Binomial Theorem
We can see that all the coefficients in Pascals
Triangle can be rewritten in the form C n .
r
The expansion of (x ? y)n can be obtained by the
following binomial theorem
12
2.2 Binomial Theorem
B. Binomial Theorem
Example 2.4T
Expand (5x ? 3)5 in ascending powers of x as far
as the 4th term.
Solution
13
2.2 Binomial Theorem
B. Binomial Theorem
Example 2.5T
Solution
(a) The 5th term
(c) General term
For the constant term, the power of x is 0.
i.e., 48 ? 4r ? 0
(b) The term in x2
r ? 12
? Constant term
14
2.2 Binomial Theorem
B. Binomial Theorem
Example 2.6T
It is given that (2 ? 3x)n ? 64 ? 18a5x ? terms
involving higher powers of x. Find the values of
n and a, where n is a positive integer.
Solution
(2 ? 3x)n
? 2n ? 64
n ? 6
15
2.2 Binomial Theorem
B. Binomial Theorem
Example 2.7T
(a) Expand (1 ? x)5(5 ? mx)3 in ascending powers
of x as far as the term in x2. (b) If the
coefficient of x2 in the expansion in (a) is 560,
find the values of m.
Solution
(a) (1 ? x)5(5 ? mx)3
(b) ? Coefficient of x2 ? 560
?
16
Chapter Summary
2.1 Summation Notation
The sum of n terms of a series with the general
term T(n) can be denoted by where n is a
positive integer.
17
2.2 Binomial Theorem
Chapter Summary
Pascals Triangle
1 1 1 1 2 1 1 3
3 1 1 4 6 4 1 1 5 10 10 5 1
? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
18
Follow-up 2.1
2.1 Summation Notation
Solution
(a)
(b)
19
Follow-up 2.2
2.1 Summation Notation
Use summation notation to express the series 1 ?
8 ? 27 ? ... ? 512.
Solution
Let T(n) be the general term of the sequence 1,
8, 27, ... , 512.
Note that

?
? 512 ? 83
?
20
Follow-up 2.3
2.2 Binomial Theorem
A. Pascals Triangle
Solution
(a) (2a ? 3b)4
(b)
21
Follow-up 2.4
2.2 Binomial Theorem
B. Binomial Theorem
Solution
22
Follow-up 2.5
2.2 Binomial Theorem
B. Binomial Theorem
Solution
(a) The 4th term
(c) General term
For the constant term, the power of x is 0.
i.e., 6 ? 2r ? 0
(b) The term in x3
r ? 3
? Constant term
23
Follow-up 2.6
2.2 Binomial Theorem
B. Binomial Theorem
It is given that (p ? qx)7 ? 128 ? 672x ? terms
involving higher powers of x. Find the values of
p and q.
Solution
(p ? qx)7
? p7 ? 128
p ? 2
24
Follow-up 2.7
2.2 Binomial Theorem
B. Binomial Theorem
(a) Expand (p ? x)3(1 ? 5x)4 in ascending powers
of x as far as the term in x2. (b) If the
constant term in the expansion in (a) is 27, find
the value of p.
Solution
(a) (p ? x)3(1 ? 5x)4
(b) ? Constant term ? 27
? p3 ? 27
p ? 3