Transcript Module 2 Chapter 2 Binomial Theorem

2
Binomial Theorem
Case Study
2.1
Summation Notation
2.2
Binomial Theorem
Chapter Summary
Case Study
You may do this, but if n
is large, it will be very
time-consuming!
Is it possible to find the value of
(1.01)n by considering the expansion
of (1  0.01)n step by step without
using a calculator?
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.
P. 2
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 ‘Σ’ 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:
n
 T (i)  T (1)  T (2)  ...  T (n)
i 1
The notation is read as ‘the summation of T(i) as i goes from 1 to n’.
n
general term of the sequence
i changes from 1 to n
 T (i)
i 1
Note:
Besides i, we can also use other letters such as j, k, m, etc, in the
n
n
k 1
m 1
summation notation such as  T ( k ) or  T ( m) .
P. 3
2.1 Summation Notation
Example 2.1T
Evaluate the following sums.
5
1
(a) 
(b)
r
r 1
7
 k (k  1)
k 3
Solution:
5
(a)
1 1 1 1 1 1
 r 1 2  3  4  5
r 1

137
60
7
(b)
 k (k  1)  3(3  1)  4(4  1)  5(5  1)  6(6  1)  7(7  1)
k 3
 110
P. 4
2.1 Summation Notation
Example 2.2T
Use summation notation to express the series
1 1 1
1
   ...  .
2 3 4
8
Solution:
Let T(n) be the general term of the sequence 1,
1 1 1
1
, , , ... , .
2 3 4
8
Note that
T ( 2) 
1
2
∴
T ( n) 
1
n
∴
1 1 1
1 8 1
    
2 3 4
8 k 2 k
T (3) 
1
3
T ( 4) 
1
4
…
T (8) 
P. 5
1
8
2.2 Binomial Theorem
A. Pascal’s 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

1
∴
(x 

x4

4x3y

6x2y2
1
3
3
1
1
3
1
1
3
1
3
3
3
1
1
3
4
6
4
1
coefficients )
1
y3
Using long multiplication again,
(x  y)4  (x  y)3(x  y)
 (x3  3x2y  3xy2  y3)(x  y)
y)4
1
1
2
1
2
1
1
2

4xy3
)

1
y4
1
P. 6
1
1
2.2 Binomial Theorem
A. Pascal’s Triangle
The expansion (x  y)n follows some special pattern.
Binomial Expansion
Coefficients of the expansion
(x  y)0
(x  y)1
(x  y)2
(x  y)3
(x  y)4
(x  y)5
1
1
1
1
1
1
2
3
4
5
1
1
3
1
6
10
4
10
1
5
1
The above triangle is called Pascal’s 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:
1
3
3
1
1
4
6
4
1
P. 7
2.2 Binomial Theorem
A. Pascal’s 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.
P. 8
2.2 Binomial Theorem
A. Pascal’s Triangle
Example 2.3T
Using the properties of Pascal’s Triangle, expand the following expressions.
(a) (2  ax)4
(b) (x2  1)5
Solution:
(a)
(2  ax)4
 1 24  4  23  (ax)1  6  22  (ax)2  4  21  (ax)3  1 (ax)4
 16  32ax  24a 2 x 2  8a3 x3  a 4 x 4
(b)
(x2  1)5
 ( x 2 )5  5( x 2 ) 4 (1)  10( x 2 )3 (1) 2  10( x 2 ) 2 (1)3
 5( x 2 )(1) 4  (1)5
 x10  5x8  10x 6 10x 4  5x 2 1
P. 9
2.2 Binomial Theorem
B. Binomial Theorem
Pascal’s 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.
n!
Recall the definition of the notation Crn: Crn 
r! (n  r )!
Rewrite Pascal’s Triangle using the notation:
P. 10
2.2 Binomial Theorem
B. Binomial Theorem
We can see that all the coefficients in Pascal’s Triangle
can be rewritten in the form Crn .
The expansion of (x  y)n can be obtained by the following binomial
theorem:
Binomial Theorem
If n is a positive integer, then
(x  y)n  C0n x n  C1n x n1 y1  C2n x n2 y 2  ...  Crn x nr y r  ...  Cnn y n

n
n nr r
C
 rx y
r 0
Using mathematical induction, we can prove the binomial theorem,
that is,
(x  y)n  C0n x n  C1n x n1 y1  C2n x n2 y 2  ...  Crn x nr y r  ...  Cnn y n
for all positive integers n.
P. 11
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:
(5x  3)5  (3  5x)5
 C05 (3) 5  C15 (3) 4 (5 x)  C 25 (3) 3 (5 x) 2  C35 (3) 2 (5 x) 3  
 1(3)5  5(3)4 (5) x  10(3)3 (5)2 x2  10(3)2 (5)3 x3  
 243  2025x  6750x2  11 250x3  
P. 12
2.2 Binomial Theorem
B. Binomial Theorem
Example 2.5T
(a) Find the 5th term in the expansion of (5  3x)6 in ascending powers of x.
(b) Find the term containing x2 in the expansion of (2  7x)5.
16
 3 1
(c) Find the constant term in the expansion of  x   .
x

Solution:
(a)
The 5th term
 C46 (5)2 (3x)4
65

 25  81x 4
2
 30 375x 4
(b) The term in x2
 C25 (2)3 (7 x)2
 10  8  49 x 2
 3920x 2
(c) General term 

C r16 ( x 3 )16  r  
1

x


r
 Cr16 (1)r  x48  4r
For the constant term, the power of x is 0.
i.e., 48  4r  0
r  12
16
∴ Constant term  C12
(1)12
 1820
P. 13
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  C0n 2 n  C1n 2 n  1 (3 x)  
 2n  3n  2n  1 x  
∴
2n  64
n 6
3n  2n  1  18a5
3(6)  26  1  18a5
a 5  25
a2
P. 14
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
 [C05 (1)5  C15 (1) 4 ( x)  C25 (1)3 ( x) 2  ]
 [C03 53  C13 52 (mx )  C23 5(mx ) 2  ]
 (1  5x  10x 2  )(125  75mx  15m2 x 2  )
 125  (75mx  625x)  (15m2 x2  375mx2  1250x2 )  
 125  (75m  625) x  (15m 2  375m  1250) x 2  
(b) ∵
∴
Coefficient of x2  560
15m 2  375m  1250  560
15m 2  375m  690  0
m 2  25m  46  0
(m  2)(m  23)  0
m  2 or 23
P. 15
Chapter Summary
2.1 Summation Notation
The sum of n terms of a series with the general term T(n)
can be denoted by
n
 T (i)  T (1)  T (2)  ...  T (n)
i 1
where n is a positive integer.
P. 16
Chapter Summary
2.2 Binomial Theorem
Pascal’s Triangle
1
1
1
2
1
1
1



3
4
5

1
1
3
6
10


4
10


1
1
5


1





Binomial Theorem
If n is a positive integer, then
(x  y)n  C0n x n  C1n x n1 y1  C2n x n2 y 2  ...  Crn x nr y r  ...  Cnn y n
n
  Crn x nr y r
r 0
P. 17