Hdhdhd 3u3q6x

3.2 BINOMIAL EXPANSION

Learning outcomes : At the end of the topic, students should be able to : (a)expand (a + b)n where n is a positive integer.

 n (b)write n! notation and Cr    as a binomial coefficient r n

(c) determine the general term in a binomial expansion (a + b)n where n is a positive integer. (d) expand (1 + x)n for

x

where n is a rational number

Expansion of (a + b)n We know that :

n=0,(a+b)0 = n=1,(a+b)1 = n=2,(a+b)2

=

1 1 a+ 1 b 1a2 + 2ab +1 b2

n=3,(a+b)3 =

1 a3+ 3a2b+ 3ab2+1 b3

n=4,(a+b)4=

1a4 +4a3b+6a2b2+4ab3+1 b4

5 5 a 1 n=5,(a+b) =

+5a4b+10a3b2+10a2b3+5ab4+1 b5

Expansion of (a + b)n 1 a+ b a2 + 2ab + b2 a3+ 3a2b+ 3ab2+ b3 a4 +4a3b+6a2b2+4ab3+b4 a5 +5a4b+10a3b2+10a2b3+5ab4+b5

The coefficient of the expansion can be arranged in a triangular array of numbers, called Pascal’s Triangle .

Extracting the coefficients alone, we obtain a table as follows n=0 1 n=1 1 ( ) 1 n=2 1 ( ) 2 ( ) 1 ( ) 1 3 ( ) 1 ( ) 3 1 4 6 4 1 This table of coefficients is known as Pascal’s triangle. Pascal triangle can be used to expand n expressions of the type (a  x)

Expansion of (a + b)n a0 1 b0 a1 1

( )

1 b1

a2 1 b0 ( ) a1 2 b1 ( ) a0 1 b2 a3 1 b0 ( ) a2 3 b1 ( ) a1 3 b2 ( ) a0 1 b3

a41 b0

a3 4 b1

a2 6 b2

a1 4 b3

(a +b)n

n=3

a0 1 b4

Pascal’s Triangle can be used to expand expressions of the type

n=0 n=1 n=2

Definition of n factorial ( n! )

+ If n  Z , then n! is given by n! = n.(n-1).(n-2).(n-3). … .3.2.1 where

0! = 1

1! = 1 Note

n! = n.(n-1)! n! = n.(n-1).(n-2)!

Example 2 a) 5! = 5.4.3.2.1 = 120 b) 2! = 2.1 = 2

c) 8!  8.7.6.5.4.3.2.1 = 8.7.6! = 56 6.5.4.3.2.1 6! 6!

Example 2

a)

Example 3 Simplify

 n +2  ! =  n +2 . n +1.n. n-1 !  n-1 !  n-1 ! =n.(n+1)(n+2)

Definition of a Binomial coefficient Let n and r be nonnegative integers, 0 ≤ r ≤ n

n n n! = C = r   r! n-r !   r  

Evaluate

5 a)   = 2

5

Example 4

C2

5! = 2! (5-2)! 5! = 2! 3! 5.4.3.2.1 = (2.1)(3.2.1)

=10

4! 4 b)   =  0  0!(4 - 0)! 4! = 0!4!

=1 n n n!   = Cr = r!  n-r  ! r 

Example 5

n  n  Show that   =   r   n-r  Solution

n!  n  RHS:   = (n-r)! n-(n-r) !    n-r  n n!  =  r  r!  n-r  !

n! = (n-r)!(n-n +r)! n! = (n -r)!r!

Continue… RHS:

LHS:

n! (n -r)!r!

compare

n n!   = r!(n -r)! r  LHS =RHS

n  n  Hence,   =    r   n-r 

BINOMIAL THEOREM For n a positive integer:  n  n 0  n  n-1 1  n  n-2 2 n 0 n (a +b) =   a b +   a b +   a b +...+   a b 0  1 2 n n

 n  n-1  n  n-r 2 n a + a b + a b +...+b (a +b) =      1 2 n

n

Notes a) The power of ‘a’ decreases by 1 for each

term from left to right. b) The power of ‘b’ increases by 1 for each

term from left to right. c) The sum of the powers of a and b always

add up to n in each . d) The expansion has (n + 1)

Example 6 Use the binomial theorem to expand the following

 n  n-1  n  n-r r (a +b) = a +   a b +   a b +...+bn  1 2 n

n

a = 4 , b = x , n = 3

 3 2 1  3 1 2 a) (4 + x) = 4 +   4 x +   4 x +x 3  1 2 3

3

= 64 + 48x +12x 2 + x 3

Example 6 (b)  n  n-1  n  n-r r n n n (a +b) = a +   a b +   a b +...+b  1 2 a = 2x , b = -y , n = 4 4 3 1 4 b) (2x - y) = (2x) +   (2x) (-y) +   (2x)2 (-y)2  1 2 4

4

4 +   (2x)1 (-y) 3 +(-y) 4  3

=16x 4 -32x 3 y +24x 2 y2 -8xy 3 + y 4

Example 7 Expand (1 + x)10 up to term in x3. Find the

value of (1.01)10 and (0.99)10 Solution

 n  n-1  n  n-r r n (a +b) = a +   a b +   a b +...+b  1 2 n

1+ x 

10

n

 10  1  10  2  10  3 =1 +   x +   x +   x +...  1 2  3 10

=1+10x + 45x 2 +120x 3 +...

Continue…

From previous expansion

10 10 10 = (1+0.01) (1.01)  (1+ x) Find

x = 0.01

(1+ x)10 =1+10x + 45x 2 +120x 3 +... 2 3 =1+10(0.01) + 45(0.01) +120(0.01) +... (1.01) 10

=1.10462

#

Continue…

From previous expansion

Find (0.99)

10

=[1+(-0.01)]  (1+ x) 10

10

x =-0.01

(1+ x)10 =1+10x + 45x 2 +120x 3 +... 2 3 =1+10(-0.01) + 45(-0.01) +120(-0.01) + (0.99)

10

= 0.90438

#

Example 8 Expand (3 - 2x)4 in the expansion ascending power of x . Hence find the value of (2.98)4 correct to 5 significant figures.

Solution

 n  n-1  n  n-r r n (a +b) = a +   a b +   a b +...+b  1 2 n

 3-2x 

n

4



a =3 , b=-2x , n = 4

Continue…

 n  n-1  n  n-r r (a +b) = a +   a b +   a b +...+bn  1 2 n

n

a =3 , b=-2x , n = 4

 3-2x 

4

4 3 1 = 3 +   3 -2x   1 4

4 2 2 +   3 -2x  2 4 1 3 +   3 -2x  + -2x  4  3

= 81+ 4  27 -2x  +6  9   4x2  + 4  3  -8x 3  +16x 4

= 81-216x +216x -96x +16x 2

3

4

Continue…

From previous expansion

4  (3-2x) (2.98)

4

Find

(3-2x) = (2.98) 2x = 0.02

x = 0.01

 3-2x  (2.98)

4

= 81-216x +216x 2 -96x 3 +16x 4

4

=81-216(0.01) +216(0.01)2 -96(0.01) 3 +16(0.01) 4

= 78.862

(5 significant figures)

Determine the general term in a Binomial Expansion, (a + b)n

The formula for the general term are r term:

 n  n-r+1 r-1 Tr =  b a  r -1 or

(r+1) term:

 n  n-r r Tr+1 =   a b r 

Example 9 Find the 6th term in the expansion of (2+3x)10 as a series in ascending power of x. r = 6 Solution

a = 2 , b = 3x , n = 10 , r = 6

 n  n-r+1 r-1 Tr =  b a  r -1

 10  10-6+1 6-1 T6 =  (3x)  (2)  6-1

Continue…

10  5  5 T6 =  2 (3x)   5 

 10  5 5 5 = 2 3 x  5 

 T6 = 1959552x 5

Example 10 Find the term that have x3 in the expansion of 5    3x +  2x  

5

Solution 5 a = 3x , b = , n =5 , r =? 2x r-1  n  n-r+1 r-1 5   5   5-r+1 Tr =  a b Tr =  (3x)     r -1    2x   r-1

Continue…

r-1

 5  6-r  5  Tr =   (3x)    2x   r -1

 5  6-r 6-r  5  =  (3) (x)   2  r -1

r-1

r-1

 5  6-r 6-r  5  =  (3) (x)   2  r -1

r-1

 5  6-r 6-r  5  =  (3) (x)   2  r -1

 5  6-r  5  =  (3)   2  r -1

r-1

(x)

r-1

 1   x

x 

-1 r-1

(x)

-r+1

-r+1+(6-r)

Continue…

 5  6-r  5  Tr =   (3)   2  r -1

r-1

(x)

7-2r

Find r to get the term that has x3

(x) 7-2r = (x) 3 7-2r = 3 r =2 2-1

 5  6-2  5  T2 =   (3)   2  2 -1

(x)

7-2(2)

Continue…

1

5 45 T2 =   (3)   (x) 3 2  1 2025 3 T2 = x 2

 The 2th term that has x3

Example 11 Find the coefficient that have x4 in the

expansion of

 2x -1

15

Solution a = 2x , b = -1 , n = 15 , r = ?

 n  n-r+1 r-1 Tr =  b a  r -1

 15  15-r+1 r-1 Tr =  (2 x) (-1 )   r -1

Continue…

 15  16-r r-1 Tr =   (2x) (-1)  r -1  15  16-r 16-r r-1 Tr =  (x) (-1)  (2)  r -1

Find r to get the term that has coefficient x4

(x)16-r = (x) 4 16-r = 4

r =12

Continue…

r =12  15  16-r 16-r r-1 Tr =  (x) (-1)  (2)  r -1

 15  16-12 16-12 12-1 T12 =  (2) (x) (-1)   12 -1  15  =   (2) 4 (x) 4 (-1)11  11  = -21840 x 4  Coefficient of x = -21840 4

#

Example 12 Find the term that is independent of x in

the expansion of

12

1  3  2x -  x  Solution

a = 2x

3

1 , b =- , n =12 , r = ? x

 n  n-r+1 r-1 Tr =  b a  r -1 r-1

 12  3 12-r+1  1  Tr =  -   (2x )   x  r-1

Continue…

 12  3 13-r r-1 -1 r-1 Tr =   (2x ) (-1) (x )  r -1

 12  13-r 3 13-r r-1 1-r =  (2) (x ) (-1) (x)  r -1  12  13-r 39-3r r-1 1-r = (2) (x) (-1) (x)   r -1  12  13-r r-1 40-4r = (2) (-1) (x)   r -1 Find r to get the term that is independent of x

Continue… Find r to get the term that is independent of x

(x)

40-4r

= (x)

0

40-4r = 0 r =10

 12  13-r r-1 40-4r Tr =  (2) (-1) (x)   r -1

 12  13-10 10-1 40-4(10) T10 =  (2) (-1) (x)   10-1

Continue… to get the term that is independent of x

 12  3 9 0 T10 =   (2) (-1) (x) 9 T10 = -1760

#

Expand ( 1 + x )n where n is a  negative integer or rational number n is positive n(n-1) 2 n  n-1 n-2  3 n = 1 +nb + b + b +...+b (1+b) 2! 3! n

n is negative or rational number

n(n-1) 2 n  n-1 n-2  3 (1+b) = 1+nb + b + b +... 2! 3! n

and the expansion valid for

-1<1

b <1

Remark

1) This expansion is an infinite series of ascending powers of x.

2) This expansion is only valid if |b| < 1

Example 13 Find the first FOUR in the expansion

of the following functions in ascending power of x and state the value of x for which the expansion are valid,

a) 1+ x 

1 4

b)

1

1+2x 

 1+2x 

-4

4

Solution (a)

(1+b)

n

n(n-1) 2 n  n-1 n-2  3 = 1+nb+ b + b +... 2! 3!

(1+ x)

1 4

 (1+b)

n



1 b=x , n= 4

 1  1  2 1 -1 ( x )    1   (1+ x) 4 =1 +   (x) +  4   4  4 2!

 1  1  1  3 -1 -2 ( x )     4 4 4   +   + ... 3!

Continue… (1+ x)

1 4

 1  3  2 ( x )    4 4  1  =1+   (x) +    2 4  1  3  7  3 ( x )     4 4  4  +   + ... 6

1 3 2 7 3 =1+ xx + x +... 4 32 128

Finally, check the validity : |b| < 1 b = x |x| < 1

 The expansion valid if

1 < x < 1

Solution (b)

(1+b)

n

n(n-1) 2 n  n-1 n-2  3 = 1+nb+ b + b +... 2! 3!

1

1+2x 

 1+2x   (1+b)n  b= 2x , n =-4 -4

4

1+2x 

-4

=1+(-4)(2x) + (-4)(-4 -1)(2x) 2!

2

(-4)(-4 -1)(-4 -2)(2x) 3 + +... 3!

Continue…

(1+2x) =1+(-4)(2x) -4

(-4)(-5)(4x ) + 2 2

(-4)(-5)(-6)(8x 3 ) + +... 3! =1-8x + 40x 2 -160x 3 +... The expansion is valid if:

b <1  2x <1

b =2x



1 1 - <x< 2 2

Expand (a + b)n where n is  negative integer or rational number

 convert it to standard form !!!

a + b

n

n

b   b  n  = a  1+   = a  1+  a a    

and the expansion valid for

b <1 a

n

Example 14 Find the first FOUR in the expansion of the following functions in ascending power of x and state the value of x for which the expansion are valid,

(a)  2 + x 

-2

(b)  8-x 

1 3

Solution (a)

2 + x 

-2

-2

  x  = 2  1+   2    =2

-2

-2

x   1+  2 

Use Binomial exp. -2

1 x =  1+  4 2

 (1+b)n x b= , n =-2 2

(1+b)

n

n(n-1) 2 n  n-1 n-2  3 = 1+nb + b + b +... 2! 3! 2

-2

x   1+   2

x -2 -2 -1    x 2 = 1+ -2    + 2! 2

3

x b= , n =-2 2

x -2 -2 -1-2 -2    2  + +... 3!

3 2 1 3 = 1- x + x - x +... 4 2

 2 + x 

-2

-2

1 x =  1+  4 2



1 3 2 1 3  2 + x   =  1- x + x - x +... -2

4

4

2



1 1 3 2 1 3 = - x+ x - x +... 4 4 16 8

The expansion valid if: -2

(refer  1+ x  )  2

x <1 2



x -1< <1 2

x b= 2

 -2< x <2

Solution (b) 1 3

  x  (8- x) = 8  1-   8   

1 3

1 3

1 3

x  = 8  1-  8  Use Binomial exp. 

x  = 2  1-  8 

1 3

 (1+b)

n

x 1 b =- , n = 8 3

(1+b)

n

n(n-1) 2 n  n-1 n-2  3 = 1+nb + b + b +... 2! 3!

 1  1  x  x  1   x   3   3 -1  - 8      1-  =1+      +    8   3  8  2! 1 3

x 1 b =- , n = 8 3

2

3

 1  1  1  x    -1 -2  -  3  3  3  8   + +... 3!

1 1 2 5 3 = 1xx x +... 24 576 41472

1 3

1 1 2 5    x 3 x x +...  2  1-  = 2  1- x576 41472  24   8

1 1 2 5 = 2x+ x x 3 +... 12 288 20736

The expansion valid if:

(refer

x - <1 8



1 3

 x )  1-   8

x -1< <1 8



x b =8

-8 < x <8

Example 15

Expand

1-2x 

1 2

in ascending powers of

x, up to the term in

x3.

in this series, evaluate decimal places.

1 By letting x = 100

2 correct to 5

Solution

1-2x  1-2x 

1 2

1 2

 1+b 

n

1  b = -2x , n = 2

1 2 1 3 =1- x- x - x +... 2 2

The expansion valid if |-2x |<1  -1<2x <1

To evaluate Subs.



|b|<1 1 1 - <x< 2 2

2

1 x= into LHS & RHS 100

LHS:

1-2x 

1 2

  1  =  1- 2    100   

 98  =   100 

1 2

 49  2  =   100  7 2 = 10

1 2

1 2

RHS:

1 2 1 3 1- x- x - x +... 2 2 2

 1  1 1  1 1  =1-  -   -    100  2  100  2  100 

3

=1-0.01-0.00005-0.0000005

= 0.9899495

Since LHS = RHS

7 2 = 0.9899495 10

10 2 = × 0.9899495 7 =1.41421

#

(5 decimal places)

This is APPROXIMATION value, to check, use calculator !