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 Page 1


 
 
 
 
240 Binomial Theorem  
  
 
 
 
 
 
 
 6.1.1 Binomial Expression. 
 An algebraic expression consisting of two terms with +ve or – ve sign between them is 
called a binomial expression.  
 For example : 
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
3 4 2
4 1
, ), 3 2 ( ), (
y x x
q
x
p
y x b a etc.  
 6.1.2 Binomial Theorem for Positive Integral Index . 
 The rule by which any power of binomial can be expanded is called the binomial theorem. 
 If n is a positive integer and x, C y ? then  
 
n
n
n n
n
n r r n
r
n n n n n n n n
y x C xy C y x C y x C y x C y x C y x
0 1
1
2 2
2
1 1
1
0 0
0
...... ........ ) ( ? ? ? ? ? ? ? ? ?
?
?
? ? ? ?
 
 i.e.,   
?
?
?
? ?
n
r
r r n
r
n n
y x C y x
0
. . ) (    .....(i)  
 Here 
n
n n n n
C C C C ,...... , ,
2 1 0
 are called binomial coefficients and 
! ) ( !
!
r n r
n
C
r
n
?
? for n r ? ? 0 . 
  
Important Tips 
? The number of terms in the expansion of 
n
y x ) ( ? are (n + 1). 
? The expansion contains decreasing power of x and increasing power of y. The sum of the powers of x and y in each 
term is equal to n. 
? The binomial coefficients ........ , ,
2 1 0
C C C
n n n
 equidistant from beginning and end are equal i.e., 
r n
n
r
n
C C
?
? . 
? ? ?
n
y x ) ( Sum of odd terms + sum of even terms.   
 
 6.1.3 Some Important Expansions . 
 (1) Replacing y  by – y  in (i), we get, 
 
?
y x C y x C y x C y x C y x C y x
n
n n r r n
r
n r n n n n n n n
. ) 1 ( .... . ) 1 ( .... . . . ) (
0 2 2
2
1 1
1
0 0
0
? ? ? ? ? ? ? ? ?
? ? ? ?
 
 i.e., 
?
?
?
? ? ?
n
r
r r n
r
n r n
y x C y x
0
. ) 1 ( ) (       .....(ii) 
 The terms in the expansion of 
n
y x ) ( ? are alternatively positive and negative, the last term 
is positive or negative according as n is even or odd. 
 (2) Replacing x by 1 and y by x in equation (i) we get,   
 
n
n
n r
r
n n n n n
x C x C x C x C x C x ? ? ? ? ? ? ? ? ...... ...... ) 1 (
2
2
1
1
0
0
 i.e., 
?
?
? ?
n
r
r
r
n n
x C x
0
) 1 ( 
 This is expansion of 
n
x) 1 ( ? in ascending power of x. 
240 
Page 2


 
 
 
 
240 Binomial Theorem  
  
 
 
 
 
 
 
 6.1.1 Binomial Expression. 
 An algebraic expression consisting of two terms with +ve or – ve sign between them is 
called a binomial expression.  
 For example : 
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
3 4 2
4 1
, ), 3 2 ( ), (
y x x
q
x
p
y x b a etc.  
 6.1.2 Binomial Theorem for Positive Integral Index . 
 The rule by which any power of binomial can be expanded is called the binomial theorem. 
 If n is a positive integer and x, C y ? then  
 
n
n
n n
n
n r r n
r
n n n n n n n n
y x C xy C y x C y x C y x C y x C y x
0 1
1
2 2
2
1 1
1
0 0
0
...... ........ ) ( ? ? ? ? ? ? ? ? ?
?
?
? ? ? ?
 
 i.e.,   
?
?
?
? ?
n
r
r r n
r
n n
y x C y x
0
. . ) (    .....(i)  
 Here 
n
n n n n
C C C C ,...... , ,
2 1 0
 are called binomial coefficients and 
! ) ( !
!
r n r
n
C
r
n
?
? for n r ? ? 0 . 
  
Important Tips 
? The number of terms in the expansion of 
n
y x ) ( ? are (n + 1). 
? The expansion contains decreasing power of x and increasing power of y. The sum of the powers of x and y in each 
term is equal to n. 
? The binomial coefficients ........ , ,
2 1 0
C C C
n n n
 equidistant from beginning and end are equal i.e., 
r n
n
r
n
C C
?
? . 
? ? ?
n
y x ) ( Sum of odd terms + sum of even terms.   
 
 6.1.3 Some Important Expansions . 
 (1) Replacing y  by – y  in (i), we get, 
 
?
y x C y x C y x C y x C y x C y x
n
n n r r n
r
n r n n n n n n n
. ) 1 ( .... . ) 1 ( .... . . . ) (
0 2 2
2
1 1
1
0 0
0
? ? ? ? ? ? ? ? ?
? ? ? ?
 
 i.e., 
?
?
?
? ? ?
n
r
r r n
r
n r n
y x C y x
0
. ) 1 ( ) (       .....(ii) 
 The terms in the expansion of 
n
y x ) ( ? are alternatively positive and negative, the last term 
is positive or negative according as n is even or odd. 
 (2) Replacing x by 1 and y by x in equation (i) we get,   
 
n
n
n r
r
n n n n n
x C x C x C x C x C x ? ? ? ? ? ? ? ? ...... ...... ) 1 (
2
2
1
1
0
0
 i.e., 
?
?
? ?
n
r
r
r
n n
x C x
0
) 1 ( 
 This is expansion of 
n
x) 1 ( ? in ascending power of x. 
240 
Binomial Theorem 241 
 (3) Replacing x by 1 and y by – x in (i) we get, 
 
n
n
n n r
r
n r n n n n
x C x C x C x C x C x ) 1 ( .... ) 1 ( ...... ) 1 (
2
2
1
1
0
0
? ? ? ? ? ? ? ? ? ?  i.e.,  
?
?
? ? ?
n
r
r
r
n r n
x C x
0
) 1 ( ) 1 ( 
 (4) .......] [ 2 ) ( ) (
4 4
4
2 2
2
0
0
? ? ? ? ? ? ?
? ?
y x C y x C y x C y x y x
n n n n n n n n
and  
 .......] [ 2 ) ( ) (
5 5
5
3 3
3
1 1
1
? ? ? ? ? ? ?
? ? ?
y x C y x C y x C y x y x
n n n n n n n n
 
 (5) The coefficient of 
th
r ) 1 ( ? term in the expansion of 
n
x) 1 ( ? is 
r
n
C . 
 (6) The coefficient of 
r
x in the expansion of 
n
x) 1 ( ? is 
r
n
C . 
 Note  : ? If n  is odd, then 
n n
y x y x ) ( ) ( ? ? ? and 
n n
y x y x ) ( ) ( ? ? ? , both have the same number of 
terms equal to .
2
1
?
?
?
?
?
? ? n
  
          ? If n is even, then 
n n
y x y x ) ( ) ( ? ? ? has ?
?
?
?
?
?
?1
2
n
 terms and 
n n
y x y x ) ( ) ( ? ? ? has 
2
n
 terms. 
 
Example: 1 ? ? ? ? ? ?
5 4 3 2 2 3 4 5
32 80 80 40 10 a xa a x a x a x x 
(a) 
5
) ( a x ? (b) 
5
) 3 ( a x ? (c) 
5
) 2 ( a x ? (d) 
3
) 2 ( a x ? 
Solution: (c) Conversely 
n
n
n n n n n n n
y x C y x C y x C C y x
0 2 2
2
1 1
1 0
.... ) ( ? ? ? ? ? ?
? ?
 
            
5
) 2 ( a x ? = 
5 0
5
5 4 1
4
5 3 2
3
5 2 3
2
5 1 4
1
5 5
0
5
) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( a x C a x C a x C a x C a x C x C ? ? ? ? ? 
                          = 
5 4 3 2 2 3 4 5
32 80 80 40 10 a xa a x a x a x x ? ? ? ? ? . 
Example: 2 The value of 
6 6
) 1 2 ( ) 1 2 ( ? ? ? will be 
 (a) – 198 (b) 198 (c) 98 (d) – 99 
Solution: (b) We know that, .....] [ 2 ) ( ) (
4 4
4
2 2
2
? ? ? ? ? ? ?
? ?
y x C y x C x y x y x
n n n n n n n
 
] ) 1 ( ) 2 ( ) 1 ( ) 2 ( ) 1 ( ) 2 ( ) 2 [( 2 ) 1 2 ( ) 1 2 (
6 0
6
6 4 2
4
6 2 4
2
6 6 6 6
C C C ? ? ? ? ? ? ? = 198 ] 1 30 4 15 8 [ 2 ? ? ? ? ? 
Example: 3 The larger of 
50 50
100 99 ? and 
50
101 is                 [IIT 1980] 
 (a) 
50 50
100 99 ? (b) Both are equal (c) 
50
101 (d) None of these 
Solution: (c) We have, ? ? ? ? ? ?
48 49 50 50 50
100
1 . 2
49 . 50
100 . 50 100 ) 1 100 ( 101 .....(i) 
 and ....... 100
1 . 2
49 . 50
100 . 50 100 ) 1 100 ( 99
48 49 50 50 50
? ? ? ? ? ?  ..... (ii) 
 Subtracting, 
50 47 50 50 50
100 ..... 100
1 . 2 . 3
48 . 49 . 50
. 2 100 99 101 ? ? ? ? ? . Hence 
50 50 50
99 100 101 ? ? . 
Example: 4 Sum of odd terms is A and sum of even terms is B in the expansion  of 
n
a x ) ( ? , then   [Rajasthan PET 1987] 
 (a) 
n n
a x a x AB
2 2
) ( ) (
4
1
? ? ? ?  (b) 
n n
a x a x AB
2 2
) ( ) ( 2 ? ? ? ? 
 (c) 
n n
a x a x AB
2 2
) ( ) ( 4 ? ? ? ?  (d) None of these 
Solution: (c) ? ? ? ? ? ? ?
? ? ? n n n
n
n n n n n n n n
a x C a x C a x C x C a x . ... ) (
2 2
2
1 1
1 0
B A a x C a x C a x C x
n n n n n n n
? ? ? ? ? ? ?
? ? ?
....) ( ..) (
3 3
3
1 1
1
2 2
2
.....(i)  
 Similarly, B A a x
n
? ? ? ) (    .....(ii) 
 From (i) and (ii), we get 
n n
a x a x AB
2 2
) ( ) ( 4 ? ? ? ?   
 Trick: Put 1 ? n in 
n
a x ) ( ? .  Then, B A a x ? ? ? . Comparing both sides a B x A ? ? , . 
Option (c) L.H.S. xa AB 4 4 ? , R.H.S. ax a x a x 4 ) ( ) (
2 2
? ? ? ? .  i.e.,  L.H.S. = R.H.S 
 
 6.1.4 General Term . 
 
n
n
n r r n
r
n n n n n n n n
y x C y x C y x C y x C y x C y x
0 2 2
2
1 1
1
0
0
.... ..... ) ( ? ? ? ? ? ? ? ?
? ? ?
 
Page 3


 
 
 
 
240 Binomial Theorem  
  
 
 
 
 
 
 
 6.1.1 Binomial Expression. 
 An algebraic expression consisting of two terms with +ve or – ve sign between them is 
called a binomial expression.  
 For example : 
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
3 4 2
4 1
, ), 3 2 ( ), (
y x x
q
x
p
y x b a etc.  
 6.1.2 Binomial Theorem for Positive Integral Index . 
 The rule by which any power of binomial can be expanded is called the binomial theorem. 
 If n is a positive integer and x, C y ? then  
 
n
n
n n
n
n r r n
r
n n n n n n n n
y x C xy C y x C y x C y x C y x C y x
0 1
1
2 2
2
1 1
1
0 0
0
...... ........ ) ( ? ? ? ? ? ? ? ? ?
?
?
? ? ? ?
 
 i.e.,   
?
?
?
? ?
n
r
r r n
r
n n
y x C y x
0
. . ) (    .....(i)  
 Here 
n
n n n n
C C C C ,...... , ,
2 1 0
 are called binomial coefficients and 
! ) ( !
!
r n r
n
C
r
n
?
? for n r ? ? 0 . 
  
Important Tips 
? The number of terms in the expansion of 
n
y x ) ( ? are (n + 1). 
? The expansion contains decreasing power of x and increasing power of y. The sum of the powers of x and y in each 
term is equal to n. 
? The binomial coefficients ........ , ,
2 1 0
C C C
n n n
 equidistant from beginning and end are equal i.e., 
r n
n
r
n
C C
?
? . 
? ? ?
n
y x ) ( Sum of odd terms + sum of even terms.   
 
 6.1.3 Some Important Expansions . 
 (1) Replacing y  by – y  in (i), we get, 
 
?
y x C y x C y x C y x C y x C y x
n
n n r r n
r
n r n n n n n n n
. ) 1 ( .... . ) 1 ( .... . . . ) (
0 2 2
2
1 1
1
0 0
0
? ? ? ? ? ? ? ? ?
? ? ? ?
 
 i.e., 
?
?
?
? ? ?
n
r
r r n
r
n r n
y x C y x
0
. ) 1 ( ) (       .....(ii) 
 The terms in the expansion of 
n
y x ) ( ? are alternatively positive and negative, the last term 
is positive or negative according as n is even or odd. 
 (2) Replacing x by 1 and y by x in equation (i) we get,   
 
n
n
n r
r
n n n n n
x C x C x C x C x C x ? ? ? ? ? ? ? ? ...... ...... ) 1 (
2
2
1
1
0
0
 i.e., 
?
?
? ?
n
r
r
r
n n
x C x
0
) 1 ( 
 This is expansion of 
n
x) 1 ( ? in ascending power of x. 
240 
Binomial Theorem 241 
 (3) Replacing x by 1 and y by – x in (i) we get, 
 
n
n
n n r
r
n r n n n n
x C x C x C x C x C x ) 1 ( .... ) 1 ( ...... ) 1 (
2
2
1
1
0
0
? ? ? ? ? ? ? ? ? ?  i.e.,  
?
?
? ? ?
n
r
r
r
n r n
x C x
0
) 1 ( ) 1 ( 
 (4) .......] [ 2 ) ( ) (
4 4
4
2 2
2
0
0
? ? ? ? ? ? ?
? ?
y x C y x C y x C y x y x
n n n n n n n n
and  
 .......] [ 2 ) ( ) (
5 5
5
3 3
3
1 1
1
? ? ? ? ? ? ?
? ? ?
y x C y x C y x C y x y x
n n n n n n n n
 
 (5) The coefficient of 
th
r ) 1 ( ? term in the expansion of 
n
x) 1 ( ? is 
r
n
C . 
 (6) The coefficient of 
r
x in the expansion of 
n
x) 1 ( ? is 
r
n
C . 
 Note  : ? If n  is odd, then 
n n
y x y x ) ( ) ( ? ? ? and 
n n
y x y x ) ( ) ( ? ? ? , both have the same number of 
terms equal to .
2
1
?
?
?
?
?
? ? n
  
          ? If n is even, then 
n n
y x y x ) ( ) ( ? ? ? has ?
?
?
?
?
?
?1
2
n
 terms and 
n n
y x y x ) ( ) ( ? ? ? has 
2
n
 terms. 
 
Example: 1 ? ? ? ? ? ?
5 4 3 2 2 3 4 5
32 80 80 40 10 a xa a x a x a x x 
(a) 
5
) ( a x ? (b) 
5
) 3 ( a x ? (c) 
5
) 2 ( a x ? (d) 
3
) 2 ( a x ? 
Solution: (c) Conversely 
n
n
n n n n n n n
y x C y x C y x C C y x
0 2 2
2
1 1
1 0
.... ) ( ? ? ? ? ? ?
? ?
 
            
5
) 2 ( a x ? = 
5 0
5
5 4 1
4
5 3 2
3
5 2 3
2
5 1 4
1
5 5
0
5
) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( a x C a x C a x C a x C a x C x C ? ? ? ? ? 
                          = 
5 4 3 2 2 3 4 5
32 80 80 40 10 a xa a x a x a x x ? ? ? ? ? . 
Example: 2 The value of 
6 6
) 1 2 ( ) 1 2 ( ? ? ? will be 
 (a) – 198 (b) 198 (c) 98 (d) – 99 
Solution: (b) We know that, .....] [ 2 ) ( ) (
4 4
4
2 2
2
? ? ? ? ? ? ?
? ?
y x C y x C x y x y x
n n n n n n n
 
] ) 1 ( ) 2 ( ) 1 ( ) 2 ( ) 1 ( ) 2 ( ) 2 [( 2 ) 1 2 ( ) 1 2 (
6 0
6
6 4 2
4
6 2 4
2
6 6 6 6
C C C ? ? ? ? ? ? ? = 198 ] 1 30 4 15 8 [ 2 ? ? ? ? ? 
Example: 3 The larger of 
50 50
100 99 ? and 
50
101 is                 [IIT 1980] 
 (a) 
50 50
100 99 ? (b) Both are equal (c) 
50
101 (d) None of these 
Solution: (c) We have, ? ? ? ? ? ?
48 49 50 50 50
100
1 . 2
49 . 50
100 . 50 100 ) 1 100 ( 101 .....(i) 
 and ....... 100
1 . 2
49 . 50
100 . 50 100 ) 1 100 ( 99
48 49 50 50 50
? ? ? ? ? ?  ..... (ii) 
 Subtracting, 
50 47 50 50 50
100 ..... 100
1 . 2 . 3
48 . 49 . 50
. 2 100 99 101 ? ? ? ? ? . Hence 
50 50 50
99 100 101 ? ? . 
Example: 4 Sum of odd terms is A and sum of even terms is B in the expansion  of 
n
a x ) ( ? , then   [Rajasthan PET 1987] 
 (a) 
n n
a x a x AB
2 2
) ( ) (
4
1
? ? ? ?  (b) 
n n
a x a x AB
2 2
) ( ) ( 2 ? ? ? ? 
 (c) 
n n
a x a x AB
2 2
) ( ) ( 4 ? ? ? ?  (d) None of these 
Solution: (c) ? ? ? ? ? ? ?
? ? ? n n n
n
n n n n n n n n
a x C a x C a x C x C a x . ... ) (
2 2
2
1 1
1 0
B A a x C a x C a x C x
n n n n n n n
? ? ? ? ? ? ?
? ? ?
....) ( ..) (
3 3
3
1 1
1
2 2
2
.....(i)  
 Similarly, B A a x
n
? ? ? ) (    .....(ii) 
 From (i) and (ii), we get 
n n
a x a x AB
2 2
) ( ) ( 4 ? ? ? ?   
 Trick: Put 1 ? n in 
n
a x ) ( ? .  Then, B A a x ? ? ? . Comparing both sides a B x A ? ? , . 
Option (c) L.H.S. xa AB 4 4 ? , R.H.S. ax a x a x 4 ) ( ) (
2 2
? ? ? ? .  i.e.,  L.H.S. = R.H.S 
 
 6.1.4 General Term . 
 
n
n
n r r n
r
n n n n n n n n
y x C y x C y x C y x C y x C y x
0 2 2
2
1 1
1
0
0
.... ..... ) ( ? ? ? ? ? ? ? ?
? ? ?
 
 
 
 
 
242 Binomial Theorem  
 The first term = 
0
0
y x C
n n
 
 The second term = 
1 1
1
y x C
n n ?
.  The third term = 
2 2
2
y x C
n n ?
 and so on 
 The term 
r r n
r
n
y x C
?
 is the 
th
r ) 1 ( ? term from beginning in the expansion of 
n
y x ) ( ? . 
 Let 
1 ? r
T denote the (r + 1)
th
 term ? 
r r n
r
n
r
y x C T
?
?
?
1
 
 This is called general term, because by giving different values to r, we can determine all 
terms of the expansion. 
 In the binomial expansion of 
r r n
r
n r
r
n
y x C T y x
?
?
? ? ? ) 1 ( , ) (
1
 
 In the binomial expansion of 
r
r
n
r
n
x C T x ? ?
?1
, ) 1 ( 
 In the binomial expansion of 
r
r
n r
r
n
x C T x ) 1 ( , ) 1 (
1
? ? ?
?
 
 Note : ? In the binomial expansion of 
n
y x ) ( ? , the p
th 
term from the end is 
th
p n ) 2 ( ? ? term 
from beginning. 
 
Important Tips 
? In the expansion of N n y x
n
? ? , ) (  
x
y
r
r n
T
T
r
r
?
?
?
?
?
? ? ?
?
?
1
1
 
? The coefficient of 
1 ? n
x in the expansion of 
2
) 1 (
) ).......( 2 )( 1 (
?
? ? ? ? ?
n n
n x x x 
? The coefficient of 
1 ? n
x in the expansion of 
2
) 1 (
) ).....( 2 )( 1 (
?
? ? ? ?
n n
n x x x 
 
Example: 5 If the 4
th
 term in the expansion of 
m
x px ) (
1 ?
? is 2.5 for all R x ? then 
 (a) 3 , 2 / 5 ? ? m p (b) 6 ,
2
1
? ? m p (c) 6 ,
2
1
? ? ? m p (d) None of these 
Solution: (b) We have 
2
5
4
? T ? 
2
5
1 3
?
?
T ? 
2
5
2
5 1
) (
6 3
3
3
3
3
? ? ? ?
?
?
?
?
?
? ? ? m m m m m
x p C
x
px C  .......(i) 
 Clearly, R.H.S. of the above equality  is independent of x  
 ? 0 6 ? ? m , 6 ? m 
 Putting 6 ? m in (i) we get ? ?
2
5
3
3
6
p C 
2
1
? p .  Hence 6 , 2 / 1 ? ? m p . 
Example: 6 If the second, third and fourth term in the expansion of 
n
a x ) ( ? are 240, 720 and 1080 respectively, 
then the value of n is  
[Kurukshetra CEE 1991; DCE 1995, 2001] 
 (a) 15 (b) 20 (c) 10 (d) 5 
Solution: (d) It is given that 1080 , 720 , 240
4 3 2
? ? ? T T T 
 Now, 240
2
? T ? 240
1 1
1 2
? ?
?
a x C T
n n
  .....(i) and  720
3
? T ? 720
2 2
2 3
? ?
?
a x C T
n n
  .....(ii) 
 1080
4
? T ? 1080
3 3
3 4
? ?
?
a x C T
n n
       .....(iii) 
 To eliminate x, 
2
1
720 . 720
1080 . 240 .
2
3
4 2
? ?
T
T T
 ? 
2
1
.
3
4
3
2
?
T
T
T
T
.  
 Now 
r
r n
C
C
T
T
r
n
r
n
r
r
1
1
1
? ?
? ?
?
?
.  Putting 3 ? r and 2 in above expression, we get  
2
1
1
2
.
3
2
?
?
?
n
n
 ? 5 ? n 
Example: 7 The 5
th
 term from the end in the expansion of 
9
3
3
2
2
?
?
?
?
?
?
?
?
?
x
x
 is  
Page 4


 
 
 
 
240 Binomial Theorem  
  
 
 
 
 
 
 
 6.1.1 Binomial Expression. 
 An algebraic expression consisting of two terms with +ve or – ve sign between them is 
called a binomial expression.  
 For example : 
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
3 4 2
4 1
, ), 3 2 ( ), (
y x x
q
x
p
y x b a etc.  
 6.1.2 Binomial Theorem for Positive Integral Index . 
 The rule by which any power of binomial can be expanded is called the binomial theorem. 
 If n is a positive integer and x, C y ? then  
 
n
n
n n
n
n r r n
r
n n n n n n n n
y x C xy C y x C y x C y x C y x C y x
0 1
1
2 2
2
1 1
1
0 0
0
...... ........ ) ( ? ? ? ? ? ? ? ? ?
?
?
? ? ? ?
 
 i.e.,   
?
?
?
? ?
n
r
r r n
r
n n
y x C y x
0
. . ) (    .....(i)  
 Here 
n
n n n n
C C C C ,...... , ,
2 1 0
 are called binomial coefficients and 
! ) ( !
!
r n r
n
C
r
n
?
? for n r ? ? 0 . 
  
Important Tips 
? The number of terms in the expansion of 
n
y x ) ( ? are (n + 1). 
? The expansion contains decreasing power of x and increasing power of y. The sum of the powers of x and y in each 
term is equal to n. 
? The binomial coefficients ........ , ,
2 1 0
C C C
n n n
 equidistant from beginning and end are equal i.e., 
r n
n
r
n
C C
?
? . 
? ? ?
n
y x ) ( Sum of odd terms + sum of even terms.   
 
 6.1.3 Some Important Expansions . 
 (1) Replacing y  by – y  in (i), we get, 
 
?
y x C y x C y x C y x C y x C y x
n
n n r r n
r
n r n n n n n n n
. ) 1 ( .... . ) 1 ( .... . . . ) (
0 2 2
2
1 1
1
0 0
0
? ? ? ? ? ? ? ? ?
? ? ? ?
 
 i.e., 
?
?
?
? ? ?
n
r
r r n
r
n r n
y x C y x
0
. ) 1 ( ) (       .....(ii) 
 The terms in the expansion of 
n
y x ) ( ? are alternatively positive and negative, the last term 
is positive or negative according as n is even or odd. 
 (2) Replacing x by 1 and y by x in equation (i) we get,   
 
n
n
n r
r
n n n n n
x C x C x C x C x C x ? ? ? ? ? ? ? ? ...... ...... ) 1 (
2
2
1
1
0
0
 i.e., 
?
?
? ?
n
r
r
r
n n
x C x
0
) 1 ( 
 This is expansion of 
n
x) 1 ( ? in ascending power of x. 
240 
Binomial Theorem 241 
 (3) Replacing x by 1 and y by – x in (i) we get, 
 
n
n
n n r
r
n r n n n n
x C x C x C x C x C x ) 1 ( .... ) 1 ( ...... ) 1 (
2
2
1
1
0
0
? ? ? ? ? ? ? ? ? ?  i.e.,  
?
?
? ? ?
n
r
r
r
n r n
x C x
0
) 1 ( ) 1 ( 
 (4) .......] [ 2 ) ( ) (
4 4
4
2 2
2
0
0
? ? ? ? ? ? ?
? ?
y x C y x C y x C y x y x
n n n n n n n n
and  
 .......] [ 2 ) ( ) (
5 5
5
3 3
3
1 1
1
? ? ? ? ? ? ?
? ? ?
y x C y x C y x C y x y x
n n n n n n n n
 
 (5) The coefficient of 
th
r ) 1 ( ? term in the expansion of 
n
x) 1 ( ? is 
r
n
C . 
 (6) The coefficient of 
r
x in the expansion of 
n
x) 1 ( ? is 
r
n
C . 
 Note  : ? If n  is odd, then 
n n
y x y x ) ( ) ( ? ? ? and 
n n
y x y x ) ( ) ( ? ? ? , both have the same number of 
terms equal to .
2
1
?
?
?
?
?
? ? n
  
          ? If n is even, then 
n n
y x y x ) ( ) ( ? ? ? has ?
?
?
?
?
?
?1
2
n
 terms and 
n n
y x y x ) ( ) ( ? ? ? has 
2
n
 terms. 
 
Example: 1 ? ? ? ? ? ?
5 4 3 2 2 3 4 5
32 80 80 40 10 a xa a x a x a x x 
(a) 
5
) ( a x ? (b) 
5
) 3 ( a x ? (c) 
5
) 2 ( a x ? (d) 
3
) 2 ( a x ? 
Solution: (c) Conversely 
n
n
n n n n n n n
y x C y x C y x C C y x
0 2 2
2
1 1
1 0
.... ) ( ? ? ? ? ? ?
? ?
 
            
5
) 2 ( a x ? = 
5 0
5
5 4 1
4
5 3 2
3
5 2 3
2
5 1 4
1
5 5
0
5
) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( a x C a x C a x C a x C a x C x C ? ? ? ? ? 
                          = 
5 4 3 2 2 3 4 5
32 80 80 40 10 a xa a x a x a x x ? ? ? ? ? . 
Example: 2 The value of 
6 6
) 1 2 ( ) 1 2 ( ? ? ? will be 
 (a) – 198 (b) 198 (c) 98 (d) – 99 
Solution: (b) We know that, .....] [ 2 ) ( ) (
4 4
4
2 2
2
? ? ? ? ? ? ?
? ?
y x C y x C x y x y x
n n n n n n n
 
] ) 1 ( ) 2 ( ) 1 ( ) 2 ( ) 1 ( ) 2 ( ) 2 [( 2 ) 1 2 ( ) 1 2 (
6 0
6
6 4 2
4
6 2 4
2
6 6 6 6
C C C ? ? ? ? ? ? ? = 198 ] 1 30 4 15 8 [ 2 ? ? ? ? ? 
Example: 3 The larger of 
50 50
100 99 ? and 
50
101 is                 [IIT 1980] 
 (a) 
50 50
100 99 ? (b) Both are equal (c) 
50
101 (d) None of these 
Solution: (c) We have, ? ? ? ? ? ?
48 49 50 50 50
100
1 . 2
49 . 50
100 . 50 100 ) 1 100 ( 101 .....(i) 
 and ....... 100
1 . 2
49 . 50
100 . 50 100 ) 1 100 ( 99
48 49 50 50 50
? ? ? ? ? ?  ..... (ii) 
 Subtracting, 
50 47 50 50 50
100 ..... 100
1 . 2 . 3
48 . 49 . 50
. 2 100 99 101 ? ? ? ? ? . Hence 
50 50 50
99 100 101 ? ? . 
Example: 4 Sum of odd terms is A and sum of even terms is B in the expansion  of 
n
a x ) ( ? , then   [Rajasthan PET 1987] 
 (a) 
n n
a x a x AB
2 2
) ( ) (
4
1
? ? ? ?  (b) 
n n
a x a x AB
2 2
) ( ) ( 2 ? ? ? ? 
 (c) 
n n
a x a x AB
2 2
) ( ) ( 4 ? ? ? ?  (d) None of these 
Solution: (c) ? ? ? ? ? ? ?
? ? ? n n n
n
n n n n n n n n
a x C a x C a x C x C a x . ... ) (
2 2
2
1 1
1 0
B A a x C a x C a x C x
n n n n n n n
? ? ? ? ? ? ?
? ? ?
....) ( ..) (
3 3
3
1 1
1
2 2
2
.....(i)  
 Similarly, B A a x
n
? ? ? ) (    .....(ii) 
 From (i) and (ii), we get 
n n
a x a x AB
2 2
) ( ) ( 4 ? ? ? ?   
 Trick: Put 1 ? n in 
n
a x ) ( ? .  Then, B A a x ? ? ? . Comparing both sides a B x A ? ? , . 
Option (c) L.H.S. xa AB 4 4 ? , R.H.S. ax a x a x 4 ) ( ) (
2 2
? ? ? ? .  i.e.,  L.H.S. = R.H.S 
 
 6.1.4 General Term . 
 
n
n
n r r n
r
n n n n n n n n
y x C y x C y x C y x C y x C y x
0 2 2
2
1 1
1
0
0
.... ..... ) ( ? ? ? ? ? ? ? ?
? ? ?
 
 
 
 
 
242 Binomial Theorem  
 The first term = 
0
0
y x C
n n
 
 The second term = 
1 1
1
y x C
n n ?
.  The third term = 
2 2
2
y x C
n n ?
 and so on 
 The term 
r r n
r
n
y x C
?
 is the 
th
r ) 1 ( ? term from beginning in the expansion of 
n
y x ) ( ? . 
 Let 
1 ? r
T denote the (r + 1)
th
 term ? 
r r n
r
n
r
y x C T
?
?
?
1
 
 This is called general term, because by giving different values to r, we can determine all 
terms of the expansion. 
 In the binomial expansion of 
r r n
r
n r
r
n
y x C T y x
?
?
? ? ? ) 1 ( , ) (
1
 
 In the binomial expansion of 
r
r
n
r
n
x C T x ? ?
?1
, ) 1 ( 
 In the binomial expansion of 
r
r
n r
r
n
x C T x ) 1 ( , ) 1 (
1
? ? ?
?
 
 Note : ? In the binomial expansion of 
n
y x ) ( ? , the p
th 
term from the end is 
th
p n ) 2 ( ? ? term 
from beginning. 
 
Important Tips 
? In the expansion of N n y x
n
? ? , ) (  
x
y
r
r n
T
T
r
r
?
?
?
?
?
? ? ?
?
?
1
1
 
? The coefficient of 
1 ? n
x in the expansion of 
2
) 1 (
) ).......( 2 )( 1 (
?
? ? ? ? ?
n n
n x x x 
? The coefficient of 
1 ? n
x in the expansion of 
2
) 1 (
) ).....( 2 )( 1 (
?
? ? ? ?
n n
n x x x 
 
Example: 5 If the 4
th
 term in the expansion of 
m
x px ) (
1 ?
? is 2.5 for all R x ? then 
 (a) 3 , 2 / 5 ? ? m p (b) 6 ,
2
1
? ? m p (c) 6 ,
2
1
? ? ? m p (d) None of these 
Solution: (b) We have 
2
5
4
? T ? 
2
5
1 3
?
?
T ? 
2
5
2
5 1
) (
6 3
3
3
3
3
? ? ? ?
?
?
?
?
?
? ? ? m m m m m
x p C
x
px C  .......(i) 
 Clearly, R.H.S. of the above equality  is independent of x  
 ? 0 6 ? ? m , 6 ? m 
 Putting 6 ? m in (i) we get ? ?
2
5
3
3
6
p C 
2
1
? p .  Hence 6 , 2 / 1 ? ? m p . 
Example: 6 If the second, third and fourth term in the expansion of 
n
a x ) ( ? are 240, 720 and 1080 respectively, 
then the value of n is  
[Kurukshetra CEE 1991; DCE 1995, 2001] 
 (a) 15 (b) 20 (c) 10 (d) 5 
Solution: (d) It is given that 1080 , 720 , 240
4 3 2
? ? ? T T T 
 Now, 240
2
? T ? 240
1 1
1 2
? ?
?
a x C T
n n
  .....(i) and  720
3
? T ? 720
2 2
2 3
? ?
?
a x C T
n n
  .....(ii) 
 1080
4
? T ? 1080
3 3
3 4
? ?
?
a x C T
n n
       .....(iii) 
 To eliminate x, 
2
1
720 . 720
1080 . 240 .
2
3
4 2
? ?
T
T T
 ? 
2
1
.
3
4
3
2
?
T
T
T
T
.  
 Now 
r
r n
C
C
T
T
r
n
r
n
r
r
1
1
1
? ?
? ?
?
?
.  Putting 3 ? r and 2 in above expression, we get  
2
1
1
2
.
3
2
?
?
?
n
n
 ? 5 ? n 
Example: 7 The 5
th
 term from the end in the expansion of 
9
3
3
2
2
?
?
?
?
?
?
?
?
?
x
x
 is  
 
 
 
 
Binomial Theorem 243 
 (a) 
3
63x (b) 
3
252
x
? (c) 
18
672
x
 (d) None of these 
Solution: (b) 5
th
 term from the end = 
th
) 2 5 9 ( ? ? term from the beginning in the expansion of 
9
3
3
2
2
?
?
?
?
?
?
?
?
?
x
x
= 
6
T  
 ? ?
6
T
3 3
4
9
5
3
4
3
5
9
1 5
252 1
. 2 .
2
2 x x
C
x
x
C T ? ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
. 
Example: 8 If 
3
2
T
T
 in the expansion of 
n
b a ) ( ? and 
4
3
T
T
 in the expansion of 
3
) (
?
?
n
b a are equal, then ? n  
[Rajasthan PET 1987, 96] 
 (a) 3 (b) 4 (c) 5 (d) 6 
Solution: (c) ? ?
?
?
?
?
?
?
?
? ?
?
a
b
n a
b
n T
T
1
2
.
1 2
2
3
2
 and ?
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
a
b
n a
b
n T
T
1
3
.
1 3 3
3
4
3
 
 ? 
4
3
3
2
T
T
T
T
?   (given) ; ? ?
?
?
?
?
?
?
? ?
?
?
?
?
?
? a
b
n a
b
n 1
3
1
2
 ? 3 3 2 2 ? ? ? n n ? 5 ? n 
 
 6.1.5 Independent Term or Constant Term. 
 Independent term or constant term of a binomial expansion is the term in which exponent of 
the variable is zero. 
 Condition : ) ( r n ? [Power of x] + r . [Power of y] = 0, in the expansion of 
n
y x ] [ ? . 
 
Example: 9 The term independent of x in the expansion of 
10
2
2
3
3
?
?
?
?
?
?
?
?
?
x
x
 will be  
[IIT 1965; BIT Ranchi 1993; Karnataka CET 2000; UPSEAT 2001] 
 (a) 
2
3
 (b) 
4
5
 (c) 
2
5
 (d) None of these 
Solution: (b) 2 0 ) 2 (
2
1
) 10 ( ? ? ? ? ? ?
?
?
?
?
?
? r r r ? 
4
5
2
3
3
1
2 2 / 8
2
10
3
? ?
?
?
?
?
?
?
?
?
?
?
?
? C T 
Example: 10 The term independent of x in the expansion of 
n
n
x
x ?
?
?
?
?
?
? ?
1
1 ) 1 ( is    [EAMCET 1989] 
 (a) 
2 2
1
2
0
) 1 ( ...... 2
n
C n C C ? ? ? ? (b) 
2
1 0
) ...... (
n
C C C ? ? ? (c) 
2 2
1
2
0
......
n
C C C ? ? ? (d) None of these 
Solution: (c) We know that, 
n
n
n n n n n
x C x C x C C x ? ? ? ? ? ? .......... ) 1 (
2
2
1
1 0
 
 
n
n
n n n n
n
x
C
x
C
x
C C
x
1
.....
1 1 1
1
2
2
1
1 0
? ? ? ? ? ?
?
?
?
?
?
? 
 Obviously, the term independent of x will be 
2 2
1
2
0 1 1 0 0
....... . ..... .
n n
n
n
n n n n n
C C C C C C C C C ? ? ? ? ? ? ? 
 Trick : Put 1 ? n in the expansion of 
x
x
x
x
x
x
1
2 1
1
1
1
1 ) 1 (
1
1
? ? ? ? ? ? ? ?
?
?
?
?
?
? ? .....(i) 
 We want coefficient of 
0
x . Comparing to equation (i). Then, we get 2 i.e., independent of x. 
 Option (c) : 
2 2
1
2
0
.....
n
C C C ? ? ;  Put 1 ? n ;  Then 2 1 1
2
1
1 2
0
1
? ? ? ? C C . 
Example: 11 The coefficient of 
7 ?
x in the expansion of 
11
2
1
?
?
?
?
?
?
?
bx
ax will be  [IIT 1967; Rajasthan PET 
1996] 
 (a) 
5
6
462
b
a
 (b) 
6
5
462
b
a
 (c) 
6
5
462
b
a ?
 (d) 
5
6
462
b
a
? 
Solution: (b) For coefficient of 
7 ?
x , 6 7 2 11 7 . ) 2 ( ) 1 ( ) 11 ( ? ? ? ? ? ? ? ? ? ? ? ? r r r r r ; 
6
5
6
5
6
11
7
462 1
) (
b
a
b
a C T ? ?
?
?
?
?
?
? ? 
Page 5


 
 
 
 
240 Binomial Theorem  
  
 
 
 
 
 
 
 6.1.1 Binomial Expression. 
 An algebraic expression consisting of two terms with +ve or – ve sign between them is 
called a binomial expression.  
 For example : 
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
3 4 2
4 1
, ), 3 2 ( ), (
y x x
q
x
p
y x b a etc.  
 6.1.2 Binomial Theorem for Positive Integral Index . 
 The rule by which any power of binomial can be expanded is called the binomial theorem. 
 If n is a positive integer and x, C y ? then  
 
n
n
n n
n
n r r n
r
n n n n n n n n
y x C xy C y x C y x C y x C y x C y x
0 1
1
2 2
2
1 1
1
0 0
0
...... ........ ) ( ? ? ? ? ? ? ? ? ?
?
?
? ? ? ?
 
 i.e.,   
?
?
?
? ?
n
r
r r n
r
n n
y x C y x
0
. . ) (    .....(i)  
 Here 
n
n n n n
C C C C ,...... , ,
2 1 0
 are called binomial coefficients and 
! ) ( !
!
r n r
n
C
r
n
?
? for n r ? ? 0 . 
  
Important Tips 
? The number of terms in the expansion of 
n
y x ) ( ? are (n + 1). 
? The expansion contains decreasing power of x and increasing power of y. The sum of the powers of x and y in each 
term is equal to n. 
? The binomial coefficients ........ , ,
2 1 0
C C C
n n n
 equidistant from beginning and end are equal i.e., 
r n
n
r
n
C C
?
? . 
? ? ?
n
y x ) ( Sum of odd terms + sum of even terms.   
 
 6.1.3 Some Important Expansions . 
 (1) Replacing y  by – y  in (i), we get, 
 
?
y x C y x C y x C y x C y x C y x
n
n n r r n
r
n r n n n n n n n
. ) 1 ( .... . ) 1 ( .... . . . ) (
0 2 2
2
1 1
1
0 0
0
? ? ? ? ? ? ? ? ?
? ? ? ?
 
 i.e., 
?
?
?
? ? ?
n
r
r r n
r
n r n
y x C y x
0
. ) 1 ( ) (       .....(ii) 
 The terms in the expansion of 
n
y x ) ( ? are alternatively positive and negative, the last term 
is positive or negative according as n is even or odd. 
 (2) Replacing x by 1 and y by x in equation (i) we get,   
 
n
n
n r
r
n n n n n
x C x C x C x C x C x ? ? ? ? ? ? ? ? ...... ...... ) 1 (
2
2
1
1
0
0
 i.e., 
?
?
? ?
n
r
r
r
n n
x C x
0
) 1 ( 
 This is expansion of 
n
x) 1 ( ? in ascending power of x. 
240 
Binomial Theorem 241 
 (3) Replacing x by 1 and y by – x in (i) we get, 
 
n
n
n n r
r
n r n n n n
x C x C x C x C x C x ) 1 ( .... ) 1 ( ...... ) 1 (
2
2
1
1
0
0
? ? ? ? ? ? ? ? ? ?  i.e.,  
?
?
? ? ?
n
r
r
r
n r n
x C x
0
) 1 ( ) 1 ( 
 (4) .......] [ 2 ) ( ) (
4 4
4
2 2
2
0
0
? ? ? ? ? ? ?
? ?
y x C y x C y x C y x y x
n n n n n n n n
and  
 .......] [ 2 ) ( ) (
5 5
5
3 3
3
1 1
1
? ? ? ? ? ? ?
? ? ?
y x C y x C y x C y x y x
n n n n n n n n
 
 (5) The coefficient of 
th
r ) 1 ( ? term in the expansion of 
n
x) 1 ( ? is 
r
n
C . 
 (6) The coefficient of 
r
x in the expansion of 
n
x) 1 ( ? is 
r
n
C . 
 Note  : ? If n  is odd, then 
n n
y x y x ) ( ) ( ? ? ? and 
n n
y x y x ) ( ) ( ? ? ? , both have the same number of 
terms equal to .
2
1
?
?
?
?
?
? ? n
  
          ? If n is even, then 
n n
y x y x ) ( ) ( ? ? ? has ?
?
?
?
?
?
?1
2
n
 terms and 
n n
y x y x ) ( ) ( ? ? ? has 
2
n
 terms. 
 
Example: 1 ? ? ? ? ? ?
5 4 3 2 2 3 4 5
32 80 80 40 10 a xa a x a x a x x 
(a) 
5
) ( a x ? (b) 
5
) 3 ( a x ? (c) 
5
) 2 ( a x ? (d) 
3
) 2 ( a x ? 
Solution: (c) Conversely 
n
n
n n n n n n n
y x C y x C y x C C y x
0 2 2
2
1 1
1 0
.... ) ( ? ? ? ? ? ?
? ?
 
            
5
) 2 ( a x ? = 
5 0
5
5 4 1
4
5 3 2
3
5 2 3
2
5 1 4
1
5 5
0
5
) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( a x C a x C a x C a x C a x C x C ? ? ? ? ? 
                          = 
5 4 3 2 2 3 4 5
32 80 80 40 10 a xa a x a x a x x ? ? ? ? ? . 
Example: 2 The value of 
6 6
) 1 2 ( ) 1 2 ( ? ? ? will be 
 (a) – 198 (b) 198 (c) 98 (d) – 99 
Solution: (b) We know that, .....] [ 2 ) ( ) (
4 4
4
2 2
2
? ? ? ? ? ? ?
? ?
y x C y x C x y x y x
n n n n n n n
 
] ) 1 ( ) 2 ( ) 1 ( ) 2 ( ) 1 ( ) 2 ( ) 2 [( 2 ) 1 2 ( ) 1 2 (
6 0
6
6 4 2
4
6 2 4
2
6 6 6 6
C C C ? ? ? ? ? ? ? = 198 ] 1 30 4 15 8 [ 2 ? ? ? ? ? 
Example: 3 The larger of 
50 50
100 99 ? and 
50
101 is                 [IIT 1980] 
 (a) 
50 50
100 99 ? (b) Both are equal (c) 
50
101 (d) None of these 
Solution: (c) We have, ? ? ? ? ? ?
48 49 50 50 50
100
1 . 2
49 . 50
100 . 50 100 ) 1 100 ( 101 .....(i) 
 and ....... 100
1 . 2
49 . 50
100 . 50 100 ) 1 100 ( 99
48 49 50 50 50
? ? ? ? ? ?  ..... (ii) 
 Subtracting, 
50 47 50 50 50
100 ..... 100
1 . 2 . 3
48 . 49 . 50
. 2 100 99 101 ? ? ? ? ? . Hence 
50 50 50
99 100 101 ? ? . 
Example: 4 Sum of odd terms is A and sum of even terms is B in the expansion  of 
n
a x ) ( ? , then   [Rajasthan PET 1987] 
 (a) 
n n
a x a x AB
2 2
) ( ) (
4
1
? ? ? ?  (b) 
n n
a x a x AB
2 2
) ( ) ( 2 ? ? ? ? 
 (c) 
n n
a x a x AB
2 2
) ( ) ( 4 ? ? ? ?  (d) None of these 
Solution: (c) ? ? ? ? ? ? ?
? ? ? n n n
n
n n n n n n n n
a x C a x C a x C x C a x . ... ) (
2 2
2
1 1
1 0
B A a x C a x C a x C x
n n n n n n n
? ? ? ? ? ? ?
? ? ?
....) ( ..) (
3 3
3
1 1
1
2 2
2
.....(i)  
 Similarly, B A a x
n
? ? ? ) (    .....(ii) 
 From (i) and (ii), we get 
n n
a x a x AB
2 2
) ( ) ( 4 ? ? ? ?   
 Trick: Put 1 ? n in 
n
a x ) ( ? .  Then, B A a x ? ? ? . Comparing both sides a B x A ? ? , . 
Option (c) L.H.S. xa AB 4 4 ? , R.H.S. ax a x a x 4 ) ( ) (
2 2
? ? ? ? .  i.e.,  L.H.S. = R.H.S 
 
 6.1.4 General Term . 
 
n
n
n r r n
r
n n n n n n n n
y x C y x C y x C y x C y x C y x
0 2 2
2
1 1
1
0
0
.... ..... ) ( ? ? ? ? ? ? ? ?
? ? ?
 
 
 
 
 
242 Binomial Theorem  
 The first term = 
0
0
y x C
n n
 
 The second term = 
1 1
1
y x C
n n ?
.  The third term = 
2 2
2
y x C
n n ?
 and so on 
 The term 
r r n
r
n
y x C
?
 is the 
th
r ) 1 ( ? term from beginning in the expansion of 
n
y x ) ( ? . 
 Let 
1 ? r
T denote the (r + 1)
th
 term ? 
r r n
r
n
r
y x C T
?
?
?
1
 
 This is called general term, because by giving different values to r, we can determine all 
terms of the expansion. 
 In the binomial expansion of 
r r n
r
n r
r
n
y x C T y x
?
?
? ? ? ) 1 ( , ) (
1
 
 In the binomial expansion of 
r
r
n
r
n
x C T x ? ?
?1
, ) 1 ( 
 In the binomial expansion of 
r
r
n r
r
n
x C T x ) 1 ( , ) 1 (
1
? ? ?
?
 
 Note : ? In the binomial expansion of 
n
y x ) ( ? , the p
th 
term from the end is 
th
p n ) 2 ( ? ? term 
from beginning. 
 
Important Tips 
? In the expansion of N n y x
n
? ? , ) (  
x
y
r
r n
T
T
r
r
?
?
?
?
?
? ? ?
?
?
1
1
 
? The coefficient of 
1 ? n
x in the expansion of 
2
) 1 (
) ).......( 2 )( 1 (
?
? ? ? ? ?
n n
n x x x 
? The coefficient of 
1 ? n
x in the expansion of 
2
) 1 (
) ).....( 2 )( 1 (
?
? ? ? ?
n n
n x x x 
 
Example: 5 If the 4
th
 term in the expansion of 
m
x px ) (
1 ?
? is 2.5 for all R x ? then 
 (a) 3 , 2 / 5 ? ? m p (b) 6 ,
2
1
? ? m p (c) 6 ,
2
1
? ? ? m p (d) None of these 
Solution: (b) We have 
2
5
4
? T ? 
2
5
1 3
?
?
T ? 
2
5
2
5 1
) (
6 3
3
3
3
3
? ? ? ?
?
?
?
?
?
? ? ? m m m m m
x p C
x
px C  .......(i) 
 Clearly, R.H.S. of the above equality  is independent of x  
 ? 0 6 ? ? m , 6 ? m 
 Putting 6 ? m in (i) we get ? ?
2
5
3
3
6
p C 
2
1
? p .  Hence 6 , 2 / 1 ? ? m p . 
Example: 6 If the second, third and fourth term in the expansion of 
n
a x ) ( ? are 240, 720 and 1080 respectively, 
then the value of n is  
[Kurukshetra CEE 1991; DCE 1995, 2001] 
 (a) 15 (b) 20 (c) 10 (d) 5 
Solution: (d) It is given that 1080 , 720 , 240
4 3 2
? ? ? T T T 
 Now, 240
2
? T ? 240
1 1
1 2
? ?
?
a x C T
n n
  .....(i) and  720
3
? T ? 720
2 2
2 3
? ?
?
a x C T
n n
  .....(ii) 
 1080
4
? T ? 1080
3 3
3 4
? ?
?
a x C T
n n
       .....(iii) 
 To eliminate x, 
2
1
720 . 720
1080 . 240 .
2
3
4 2
? ?
T
T T
 ? 
2
1
.
3
4
3
2
?
T
T
T
T
.  
 Now 
r
r n
C
C
T
T
r
n
r
n
r
r
1
1
1
? ?
? ?
?
?
.  Putting 3 ? r and 2 in above expression, we get  
2
1
1
2
.
3
2
?
?
?
n
n
 ? 5 ? n 
Example: 7 The 5
th
 term from the end in the expansion of 
9
3
3
2
2
?
?
?
?
?
?
?
?
?
x
x
 is  
 
 
 
 
Binomial Theorem 243 
 (a) 
3
63x (b) 
3
252
x
? (c) 
18
672
x
 (d) None of these 
Solution: (b) 5
th
 term from the end = 
th
) 2 5 9 ( ? ? term from the beginning in the expansion of 
9
3
3
2
2
?
?
?
?
?
?
?
?
?
x
x
= 
6
T  
 ? ?
6
T
3 3
4
9
5
3
4
3
5
9
1 5
252 1
. 2 .
2
2 x x
C
x
x
C T ? ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
. 
Example: 8 If 
3
2
T
T
 in the expansion of 
n
b a ) ( ? and 
4
3
T
T
 in the expansion of 
3
) (
?
?
n
b a are equal, then ? n  
[Rajasthan PET 1987, 96] 
 (a) 3 (b) 4 (c) 5 (d) 6 
Solution: (c) ? ?
?
?
?
?
?
?
?
? ?
?
a
b
n a
b
n T
T
1
2
.
1 2
2
3
2
 and ?
?
?
?
?
?
?
? ?
?
?
?
?
?
? ? ?
?
a
b
n a
b
n T
T
1
3
.
1 3 3
3
4
3
 
 ? 
4
3
3
2
T
T
T
T
?   (given) ; ? ?
?
?
?
?
?
?
? ?
?
?
?
?
?
? a
b
n a
b
n 1
3
1
2
 ? 3 3 2 2 ? ? ? n n ? 5 ? n 
 
 6.1.5 Independent Term or Constant Term. 
 Independent term or constant term of a binomial expansion is the term in which exponent of 
the variable is zero. 
 Condition : ) ( r n ? [Power of x] + r . [Power of y] = 0, in the expansion of 
n
y x ] [ ? . 
 
Example: 9 The term independent of x in the expansion of 
10
2
2
3
3
?
?
?
?
?
?
?
?
?
x
x
 will be  
[IIT 1965; BIT Ranchi 1993; Karnataka CET 2000; UPSEAT 2001] 
 (a) 
2
3
 (b) 
4
5
 (c) 
2
5
 (d) None of these 
Solution: (b) 2 0 ) 2 (
2
1
) 10 ( ? ? ? ? ? ?
?
?
?
?
?
? r r r ? 
4
5
2
3
3
1
2 2 / 8
2
10
3
? ?
?
?
?
?
?
?
?
?
?
?
?
? C T 
Example: 10 The term independent of x in the expansion of 
n
n
x
x ?
?
?
?
?
?
? ?
1
1 ) 1 ( is    [EAMCET 1989] 
 (a) 
2 2
1
2
0
) 1 ( ...... 2
n
C n C C ? ? ? ? (b) 
2
1 0
) ...... (
n
C C C ? ? ? (c) 
2 2
1
2
0
......
n
C C C ? ? ? (d) None of these 
Solution: (c) We know that, 
n
n
n n n n n
x C x C x C C x ? ? ? ? ? ? .......... ) 1 (
2
2
1
1 0
 
 
n
n
n n n n
n
x
C
x
C
x
C C
x
1
.....
1 1 1
1
2
2
1
1 0
? ? ? ? ? ?
?
?
?
?
?
? 
 Obviously, the term independent of x will be 
2 2
1
2
0 1 1 0 0
....... . ..... .
n n
n
n
n n n n n
C C C C C C C C C ? ? ? ? ? ? ? 
 Trick : Put 1 ? n in the expansion of 
x
x
x
x
x
x
1
2 1
1
1
1
1 ) 1 (
1
1
? ? ? ? ? ? ? ?
?
?
?
?
?
? ? .....(i) 
 We want coefficient of 
0
x . Comparing to equation (i). Then, we get 2 i.e., independent of x. 
 Option (c) : 
2 2
1
2
0
.....
n
C C C ? ? ;  Put 1 ? n ;  Then 2 1 1
2
1
1 2
0
1
? ? ? ? C C . 
Example: 11 The coefficient of 
7 ?
x in the expansion of 
11
2
1
?
?
?
?
?
?
?
bx
ax will be  [IIT 1967; Rajasthan PET 
1996] 
 (a) 
5
6
462
b
a
 (b) 
6
5
462
b
a
 (c) 
6
5
462
b
a ?
 (d) 
5
6
462
b
a
? 
Solution: (b) For coefficient of 
7 ?
x , 6 7 2 11 7 . ) 2 ( ) 1 ( ) 11 ( ? ? ? ? ? ? ? ? ? ? ? ? r r r r r ; 
6
5
6
5
6
11
7
462 1
) (
b
a
b
a C T ? ?
?
?
?
?
?
? ? 
 
 
 
 
244 Binomial Theorem  
Example: 12 If the coefficients of second, third and fourth term in the expansion of 
n
x
2
) 1 ( ? are in A.P., then 
7 9 2
2
? ? n n is equal to  
         [AMU 2001] 
 (a) –1  (b) 0 (c) 1 (d) 3/2 
Solution: (b) 
1
2
2
C T
n
? , 
3
2
4 2
2
3
, C T C T
n n
? ? are in A.P. then,  
3
2
1
2
2
2
. 2 C C C
n n n
? ? 
 
1 . 2 . 3
) 2 2 )( 1 2 ( 2
1
2
1 . 2
) 1 2 (. 2
. 2
? ?
? ?
? n n n n n n
 
 On solving, 0 7 9 2
2
? ? ? n n 
Example: 13   The coefficient of 
5
x in the expansion of 
4 5 2
) 1 ( ) 1 ( x x ? ? is                 [EAMCET 1996; 
UPSEAT 2001; Pb. CET 2002] 
 (a) 30 (b) 60 (c) 40 (d) None of these 
Solution: (b) We have 
4 5 2
) 1 ( ) 1 ( x x ? ? = .....) (
4
2
5 2
1
5
0
5
? ? ? x C x C C .......) (
2
2
4 1
1
4
0
4
? ? ? x C x C C 
So coefficient of 
5
x in ] ) 1 ( ) 1 [(
4 5 2
x x ? ? = 60 . .
1
5
3
4
1
4
2
5
? ? C C C C 
Example: 14  If A and B are the coefficient of 
n
x in the expansions of 
n
x
2
) 1 ( ? and 
1 2
) 1 (
?
?
n
x respectively, then[MP PET 1999] 
 (a) B A ? (b) B A 2 ? (c) B A ? 2 (d)  None of these 
Solution: (b) ? A coefficient of 
n
x in 
n
x
2
) 1 ( ? = 
! !
! ) 2 (
2
n n
n
C
n
n
? = 
! )! 1 (
)! 1 2 .( 2
n n
n
?
?
 .....(i) 
 ? B coefficient of 
n
x in 
)! 1 ( !
)! 1 2 (
) 1 (
1 2 1 2
?
?
? ? ?
? ?
n n
n
C x
n
n n
   .....(ii)  
 By (i) and (ii) we get, B A 2 ? 
Example: 15 The coefficient of 
n
x in the expansion of 
n
x x ) 1 )( 1 ( ? ? is      [AIEEE 2004] 
 (a) n
n 1
) 1 (
?
? (b) ) 1 ( ) 1 ( n
n
? ? (c) 
2 1
) 1 ( ) 1 ( ? ?
?
n
n
 (d) ) 1 ( ? n 
Solution: (b) Coefficient of 
n
x  in 
n
x x ) 1 ( ) 1 ( ? ? = Coefficient of 
n
x in ? ?
n
x) 1 ( coefficient of 
n n
x x ) 1 ( in
1
?
?
 
= Coefficient of 
n
x in ] ) ( . ) ( [
1
1
?
?
? ? ?
n
n
n n
n
n
x C x x C = 
1
1
. ) 1 ( ) 1 ( C C
n n
n
n n ?
? ? ? = ] 1 [ ) 1 ( ) .( ) 1 ( ) 1 ( n n
n n n
? ? ? ? ? ? ? . 
 
 6.1.6 Number of Terms in the Expansion of (a + b + c)
n
 and (a + b + c + d)
n
 . 
 
n
c b a ) ( ? ? can be expanded as : 
n n
c b a c b a } ) {( ) ( ? ? ? ? ?
 
n
n
n n n n n n
c C c b a C c b a C b a ? ? ? ? ? ? ? ?
? ?
..... ) ( ) ( ) ( ) ( ) (
2 2
2
1 1
1
term 1 ... term ) 1 (  term term ) 1 ( ? ? ? ? ? ? ? n n n 
 ? Total number of terms = 
2
) 2 )( 1 (
1 ...... ) 1 ( ) ( ) 1 (
? ?
? ? ? ? ? ? ?
n n
n n n . 
 Similarly, Number of terms in the expansion of 
6
) 3 )( 2 )( 1 (
) (
? ? ?
? ? ? ?
n n n
d c b a
n
. 
 
Example: 16  If the number of terms in the expansion of 
n
z y x ) 3 2 ( ? ? is  45, then ? n 
 (a) 7 (b) 8 (c) 9 (d)  None of these 
Solution: (b) Given, total number of terms = 45
2
) 2 ( ) 1 (
?
? ? n n
 ? 90 ) 2 )( 1 ( ? ? ? n n ? 8 ? n . 
Example: 17 The number of terms in the expansion of 
3 2 2
] ) 3 ( ) 3 [( y x y x ? ? is    [Rajasthan  PET 1986] 
 (a) 14 (b) 28 (c) 32 (d) 56 
Solution: (b) We have 
6
)] 3 )( 3 [( y x y x ? ? = 
6 2 2
] 3 8 3 [ y xy x ? ? ; Number of terms = 28
2
) 2 6 ( ) 1 6 (
?
? ?
 
 6.1.7 Middle Term. 
 The middle term depends upon the value of n.

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FAQs on Detailed Notes: Binomial Theorem - Mathematics (Maths) for JEE Main & Advanced

1. What is the Binomial Theorem and why is it important in JEE?

Ans. The Binomial Theorem provides a formula for expanding expressions of the form \((a + b)^n\), where \(n\) is a non-negative integer. It states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\), where \(\binom{n}{k}\) is a binomial coefficient. This theorem is crucial for JEE as it helps in simplifying expressions, solving combinatorial problems, and understanding polynomial expansions, all of which are common in various topics of physics and mathematics.

2. How do you find the coefficients of specific terms in a binomial expansion?

Ans. To find the coefficient of a specific term in the expansion of \((a + b)^n\), you can use the binomial coefficient \(\binom{n}{k}\), where \(k\) is the exponent of \(b\) in that term. The general term in the expansion is given by \(T_k = \binom{n}{k} a^{n-k} b^k\). For example, to find the coefficient of \(b^3\) in \((x + 2)^5\), you would calculate \(T_3 = \binom{5}{3} x^{5-3} 2^3\).

3. Can the Binomial Theorem be used for negative integers or non-integer values of \(n\)?

Ans. The Binomial Theorem as stated is applicable for non-negative integers \(n\). However, it can be generalized for any real number \(n\) using the Binomial Series, which states that \((1 + x)^n\) can be expanded as \(\sum_{k=0}^{\infty} \binom{n}{k} x^k\) for \(|x| < 1\). This series is particularly useful in calculus and for evaluating limits.

4. What is the significance of binomial coefficients in combinatorics?

Ans. Binomial coefficients \(\binom{n}{k}\) represent the number of ways to choose \(k\) elements from a set of \(n\) elements without regard to the order of selection. This concept is fundamental in combinatorics, as it helps in solving problems related to combinations, permutations, and probability. In JEE, understanding how to compute and apply these coefficients is essential for tackling combinatorial problems.

5. How can the Binomial Theorem be applied to solve problems involving probability?

Ans. The Binomial Theorem can be used in probability to model situations where there are two possible outcomes, such as success and failure. For example, in a binomial distribution, if the probability of success is \(p\) and that of failure is \(q = 1 - p\), the theorem allows us to calculate the probability of obtaining exactly \(k\) successes in \(n\) trials using the formula \(P(X = k) = \binom{n}{k} p^k q^{n-k}\). This is often tested in JEE to evaluate students' understanding of probability concepts.

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