Applied Math 62 Binomial Theorem Chapter 3 Binomial Theorem 3.1 Introduction: An algebraic expression containing two terms is called a binomial expression, Bi means two and nom means term. Thus the general type of a binomial is a + b , x – 2 , 3x + 4 etc. The expression of a binomial raised to a small positive power can be solved by ordinary multiplication , but for large power the actual multiplication is laborious and for fractional power actual multiplication is not possible. By means of binomial theorem, this work reduced to a shorter form. This theorem was first established by Sir Isaac Newton. 3.2 Factorial of a Positive Integer: If n is a positive integer, then the factorial of ‘n’ denoted by n! or n and is defined as the product of n +ve integers from n to 1 (or 1 to n ) i.e., n! = n(n – 1)(n – 2) ….. 3.2.1 For example, 4! = 4.3.2.1 = 24 and 6! = 6.5.4.3.2.1 = 720 one important relationship concerning factorials is that (n + 1)! = (n + 1) n! _________________ (1) for instance, 5! = 5.4.3.2.1 = 5(4.3.2.1) 5! = 5.4! Obviously, 1! = 1 and this permits to define from equation (1) (n+1)! n!= n+1 Substitute 0 for n, we obtain (0+1)! 1! 1 0!= 0+1 1 1 0! = 1 3.3 Combination: Each of the groups or selections which can be made out of a given number of things by taking some or all of them at a time is called combination. In combination the order in which things occur is not considered e.g.; combination of a, b, c taken two at a time are ab, bc, ca.
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Applied Math 62 Binomial Theorem
Chapter 3
Binomial Theorem
3.1 Introduction: An algebraic expression containing two terms is called a binomial
expression, Bi means two and nom means term. Thus the general type of
a binomial is a + b , x – 2 , 3x + 4 etc. The expression of a binomial
raised to a small positive power can be solved by ordinary multiplication ,
but for large power the actual multiplication is laborious and for fractional
power actual multiplication is not possible. By means of binomial
theorem, this work reduced to a shorter form. This theorem was first
established by Sir Isaac Newton.
3.2 Factorial of a Positive Integer: If n is a positive integer, then the factorial of ‘n’ denoted by n! or
n and is defined as the product of n +ve integers from n to 1 (or 1 to n )
i.e., n! = n(n – 1)(n – 2) ….. 3.2.1
For example,
4! = 4.3.2.1 = 24
and 6! = 6.5.4.3.2.1 = 720
one important relationship concerning factorials is that
(n + 1)! = (n + 1) n! _________________ (1)
for instance,
5! = 5.4.3.2.1
= 5(4.3.2.1)
5! = 5.4!
Obviously, 1! = 1 and this permits to define from equation (1)
(n+1)!
n!=n+1
Substitute 0 for n, we obtain
(0+1)! 1! 1
0!=0+1 1 1
0! = 1
3.3 Combination: Each of the groups or selections which can be made out of a given
number of things by taking some or all of them at a time is called
combination.
In combination the order in which things occur is not considered
e.g.; combination of a, b, c taken two at a time are ab, bc, ca.
Applied Math 63 Binomial Theorem
The numbers n
r
or n
cr
The numbers of the combination of n different objects taken ، r
٫ at a
time is denoted by n
r
or n
cr and is defined as,
n
r
=
( )
e.g, ( )
( )
=
Example 1: Expand ( )
Solution. ( ) =
( )
= 7.6.5.4!
3.2.1.4!
= 35
This can also be expand as
3
7.6.57
3.2.1 = 35
If we want to expand ( ) , then
5
7.6.5.4.37
5.4.3.2.1 = 21
Procedure: Expand the above number as the lower number and the lower
number expand till 1.
Method 2
For expansion of n
r
we can apply the method:
a. If r is less than (n – r) then take r factors in the numerator from n to
downward and r factors in the denominator ending to 1.
Applied Math 64 Binomial Theorem
b. If n – r is less than r, then take (n – r) factors in the numerator from
n to downward and take (n – r) factors in the denominator ending
to 1. For example, to expand 57 again, here 7 – 5 = 2 is less than
5, so take two factors in numerator and two in the denominator as,
5
7.67
2.1 = 21
Some Important Results
(i). ( ) =
( )
(ii) ( )
( )
(iii) n n
r n - r
For example
0 44 4 = 1 as
4! 4! =
0!(4-0)! 4! 0!
1 = 1
( ) (
) = 4 as
4 = 4
Note: The numbers n
r
or n
cr are also called binomial co-efficients
3.4 The Binomial Theorem: The rule or formula for expansion of (a + b)
n, where n is
any positive integral power , is called binomial theorem .
For any positive integral n
( ) ( ) (
) (
) (
)
+( ) (
) ---------------------(1)
Applied Math 65 Binomial Theorem
( ) ∑ ( )
Remarks:- The coefficients of the successive terms are
( ) (
) (
) (
) (
)
and are called Binomial coefficients.
Note : Sum of binomial coefficients is 2n
Another form of the Binomial theorem:
(a + b)n = a
n+
n
1!a
n–1b +
n-2 2n(n - 1)a b
2!+
n-3 3n(n - 1)(n - 2)a b
3!+ ……+
n-r rn(n - 1)(n - 2) - - - - - - (n - r + 1)a b
r! + ………+ b
n------------------ (2)
Note: Since,
n !
r ! !
n
r n r
So, ( ) =
( )
n n! n(n - 1)! n =
1 1! n 1 ! 1! n - 1 ! 1!
n n! n(n - 2)! n(n - 1) =
2 2! n 2 ! 2! n - 2 ! 2!
n n! n(n - 1)(n - 2)(n - 3)! n(n - 1)(n - 2) =
3 3! n 3 ! 3! n - 3 ! 3!
------------------------------
------------------------------
n n(n - 1)(n - 2)!......(n - r + 1)(n - r)!
r r! n - r !
= n(n - 1)(n - 2)......(n - r + 1)
r!
(
)
( )
Applied Math 66 Binomial Theorem
The following points can be observed in the expansion of (a + b)n
1. There are (n + 1) terms in the expansion.
2. The 1st term is a
n and (n + 1)th term or the last term is
b
n
3. The exponent of ‘a’ decreases from n to zero.
4. The exponent of ‘b’ increases from zero to n.
5. The sum of the exponents of a and b in any term is equal to index n.
6. The co-efficients of the term equidistant from the beginning and end
of the expansion are equal as n n
r n - r
3.5 General Term:
The term n - r rn
a br
in the expansion of binomial theorem is
called the General term or (r + 1)th term. It is denoted by Tr + 1. Hence
Tr + 1 = n - r rn
a br
Note: The General term is used to find out the specified term or
the required co-efficient of the term in the binomial expansion Example 2: Expand (x + y)
4 by binomial theorem:
Solution:
(x + y)4
= 4 4 4 - 1 4 4 - 2 2 4 4 - 3 3 41 2 3x + x y + x y + x y + y
= 4 3 2 2 3 44 x 3 4 x 3 x 2x + 4x y + x y + xy + y
2 x 1 3 x 2 x 1
= 4 3 2 2 3 4x + 4x y + 6x y + 4xy + y
Example 3: Expand by binomial theorem
61
a - a
Solution:
61
a - a
= 1 2 3
6 6 6 - 1 6 6 - 2 6 6 - 3
1 2 3
1 1 1a + a a a
a a a
4 5 6
6 6 - 4 6 6 - 5 6 6 - 6
4 5 6
1 1 1 a a a
a a a
= 6 5 4 3
2 3
1 6 x 5 1 6 x 5 x 4 1a + 6a a a
a 2 x 1 a 3 x 2 x 1 a
Applied Math 67 Binomial Theorem
5
2
4 5 6
6 x 5 x 4 x 3 1 6 x 5 x 4 x 3 x 2 1 1 a a
4 x 3 x 2 x 1 a 5 x 4 x 3 x 2 x 1 a a
= 6 4 2
2 5 6
15 6 1a - 6a + 15a - 20 + - +
a a a
Example 4: Expand
42x 2
- 2 x
Solution:
42x 2
- 2 x
=
4 4 - 1 4 - 21 22 2 24 4x x -2 x -2 + +
1 22 2 x 2 x
4 - 3 4 - 43 42 24 4x -2 x -2+ +
3 42 x 2 x
=
3 24 2 2
2
x x 2 4 . 3 x 44
16 2 x 2 . 1 2 x
2
3 4
4 . 3 . 2 x 8 16
3 . 2 . 1 2 x x
= 8 8 4 2
2 3 4
x x 2 x 4 x 8 164. . 6. . 4 .
16 8 x 4 x 2 x x
= 8
5 2
4
x 16 16x 6x
16 x x
Example 5: Expand (1.04)5 by the binomial formula and find its value
to two decimal places.
Solution:
(1.04)5 = (1 + 0.04)
5
(1 + 0.04)5 =
5 5 1 5 2 25 5 51 1 0.04 1 0.04
1 2 3
5 3 3 5 4 4 55
1 0.04 1 0.04 0.044
= 1. + 0.2 + 0.016 + 0.00064 + 0.000128
+ 0.000 000 1024
= 1.22
Example 6: Find the eighth term in the expansion of
12
2
2
12x
x
Applied Math 68 Binomial Theorem
Solution:
12
2
2
12x
x
The General term is, Tr + 1 = n - r rn
a br
Here T8 = ? a = 2x2, b =
2
1
x , n = 12 , r = 7 ,
Therefore , T7+1 = 7
12 72
2
12 12x
7 x
T8 = 7
52
14
12 . 11 . 10 . 9 . 8 . 7 . 6 ( 1)2x
7 . 6 . 5 . 4 . 3 . 2 . 1 x
T8 = 10
14
( 1)793 x 32x
x
T8 = 4
25 344
x
Eighth term = T8 = 4
25 344
x
.
3.6 Middle Term in the Expansion (a + b)n
In the expansion of (a + b)n, there are (n + 1) terms.
Case I :
If n is even then (n + 1) will be odd, so n
+ 12
th term will be
the only one middle term in the expension .
For example, if n = 8 (even), number of terms will be 9 (odd),
therefore, 8
+ 12
= 5th
will be middle term.
Case II:
If n is odd then (n + 1) will be even, in this case there will not be a
single middle term, but n + 1
2
th and n + 1
+ 12
th term will be the two
middle terms in the expension.
For example, for n = 9 (odd), number of terms is 10 i.e. 9 + 1
2
th and 9 + 1
12
th i.e. 5
th and 6
th terms are taken as middle terms and
these middle terms are found by using the formula for the general term.
Applied Math 69 Binomial Theorem
Example 7: Find the middle term of
142x
12
.
Solution:
We have n = 14, then number of terms is 15.
14
12
i.e. 8
th will be middle term.
a = 1, b =
2x
2 , n = 14, r = 7, T8 = ?
Tr + 1 = n - r rn
a br
T7 + 1 = 7
72 14
14 - 7 714 x 14! x1 1
2 7!7!7 2
T8 = 1414.13.12.11.10.9.8.7! (-1)
.x7.6.5.4.3.2.1 7! 128
T8 = - (2) (13) (11) (2) (3) 1
128. 14x
T8 =
Example 8 : Find the coefficient of x19
in (2x3 – 3x)
9.
Solution:
Here, a = 2x3, b = –3x, n = 9
First we find r.
Since Tr + 1 = n - r rn
a br
= 9 - r r39
2x 3xr
= r9 - r 27 - 3r r9
2 3 x .xr
= r9 - r 27 - 2r9
2 3 .xr
……………. (1)
But we require x19
, so put
19 = 27 – 2r
2r = 8
r = 4
Applied Math 70 Binomial Theorem
Putting the value of r in equation (1)
T4 + 1 = 49 - 4 19
92 3 x
r
= 5 4 199 . 8 . 7 . 6 . 5 . 2 . 3 x
4 . 3 . 2 . 1
= 630 x 32 x 81 x19
T5 = 1632960 x19
Hence the coefficient of x19
is 1632960
Example 9: Find the term independent of x in the expansion of
9
2 12x
x
.
Solution:
Let Tr + 1 be the term independent of x.
We have a = 2x2, b =
1
x, n = 9
Tr + 1 = n - r rn
a br
= r
9 - r29 1
2xxr
Tr + 1 = 9 - r 18 - 2r r92 . x . x
r
Tr + 1 = 9 - r 18 - 3r92 . x
r
…………… (1)
Since Tr + 1 is the term independent of x i.e. x0.
power of x must be zero.
i.e. 18 – 3r = 0 r = 6
put in (1)
Tr + 1 = 3
9 - 6 0
2
9 !92 . x = .1
6 !6!3
= 3 9 . 8
4 . 7.6!
6! . 3 . 2 . 8 . 1
. 1 = 672
Applied Math 71 Binomial Theorem
Exercise 3.1 1. Expand the following by the binomial formula.
(i)
41
x + x
(ii)
52x 3
- 3 2x
(iii)
4x 2
2 y
(iv) 5(2x - y) (v)
7x
2a - a
(vi) (
)
(vii) ( - x + y
-1 )
4
2. Compute to two decimal places of decimal by use of binomial
formula.
(i) (1.02)4 (ii)
60.98 (iii)
52.03
3. Find the value of
(i) 5 5(x + y) + (x - y) (ii)
4 4(x + 2) + (x - 2)
4. Expanding the following in ascending powers of x
(i) 2 4(1 x + x ) (ii)
2 4(2 x x )
5. Find
(i) the 5th
term in the expansion of
102 3
2xx
(ii) the 6th
term in the expansion of
152 y
x2
(iii) the 8th
term in the expansion of
122
xx
(iv) the 7th
term in the expansion of
94x 5
5 2x
6. Find the middle term of the following expansions
(i)
102 1
3x2x
(ii)
11a b
2 3
(iii)
71
2x+x
7. Find the specified term in the expansion of
(i)
102 3
2xx
: term involving x
5
Applied Math 72 Binomial Theorem
(ii)
102 1
2x2x
: term involving x
5
(iii)
73 1
xx
: term involving x
9
(iv)
8x 4
2 x
: term involving x
2
(v)
122
2p6q
2
: term involving q
8
8. Find the coefficient of
(i) x5 in the expansion of
102 3
2xx
(ii) x20
in the expansion of
162 1
2x2x
(iii) x5 in the expansion of
102 1
2x3x
(iv) b6 in the expansion of
102
2a2b
2
9. Find the constant term in the expansion of
(i)
92 1
xx
(ii)
10
2
1x
3x
10. Find the term independent of x in the expansion of the following
(i)
122 1
2xx
(ii)
92 1
2xx
Answers 3.1
1. (i) 4 2
2 4
4 1x + 4x + 6 + +
x x
(ii) 5 3
3 5
32 40 20 15 135 243x x x +
243 27 3 x 8x 32x
(iii)
4 3 2
2 3 4
x x 6x 6x 16 +
16 y y y y
Applied Math 73 Binomial Theorem
(iv) 5 4 3 2 2 3 4 532x 80x y + 80x y - 40x y + 10xy - y