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Binomial Theorem
8.Binomial Theorem
Solution 4:
By using Binomial Theorem, the expression
5 1
3
x
x
can be expanded as
5 5 4 3 2 2 3 4 5 5 5 5 5 5 5 0 1 2 3 4 5
1 1 1 1 1 1
3 3 3 3 3 3
x x x x x x C C C C C C x x x x x x
^ ^ ^ ^ ^ ^
5 4 3 2
2 3 4 5
1 1 1 1 1 5 10 10 5 243 81 27 9 3
x x x x x
x x x x x
(^) (^) (^) (^) (^) (^)
^ ^ ^ ^ ^ ^ ^ ^
5 3
3 5
5 10 5 1
243 81 9 3
x x x
x x x
Question 5:
Expand
6 1 x x
Solution 5:
By using Binomial Theorem, the expression
6 1 x x
can be expanded as
6 2 3 4 5 6 6 6 6 5 6 4 6 3 6 2 6 6 0 1 2 3 4 5 6
1 1 1 1 1 1 1 x C x C x C x C x C x C x C x x x x x x x
^ ^ ^ ^ ^ ^ ^
6 5 4 3 2 2 3 4 5 6
1 1 1 1 1 1 x 6 x 15 x 20 x 15 x 6 x x x x x x x
6 4 2 2 4 6
15 6 1 x 6 x 15 x 20 x x x
Question 6:
Using Binomial Theorem, evaluate
3 96
Solution 6:
96 can be expressed as the sum or difference of two numbers whose powers are easier to
calculate and then, binomial theorem can be applied.
It can be written that, 96 100 4
3 3 96 100 4
3 3 3 2 3 2 3 C 0 (^) 100 C 1 (^) 100 4 C 2 (^) 100 4 3 C 3 4
3 2 2 3 100 3 100 4 3 100 4 4
Binomial Theorem
8.Binomial Theorem
Question 7:
Using Binomial Theorem, evaluate
5 102
Solution 7:
102 can be expressed as the sum or difference of two numbers whose powers are easier to
calculate and then, binomial theorem can be applied.
It can be written that, 102 100 2
5 5 102 100 2
5 5 5 4 5 3 2 5 2 3 5 4 5 5 C 0 (^) 100 C 1 (^) 100 2 C 2 (^) 100 2 C 3 (^) 100 2 C 4 (^) 100 2 C 5 2
Question 8:
Using Binomial Theorem, evaluate
4 101
Solution 8:
101 can be expressed as the sum or difference of two numbers whose powers are easier to
calculate and then, binomial theorem can be applied.
It can be written that, 101 100 1
4 4 101 100 1
4 4 4 3 4 2 2 4 3 4 4 C 0 (^) 100 C 1 (^) 100 1 C 2 (^) 100 1 C 3 (^) 100 1 C 4 1
4 3 2 4 100 4 100 6 100 4 100 1
Question 9:
Using Binomial Theorem, evaluate
5 99
Solution 9:
99 can be written as the sum or difference of two numbers whose powers are easier to calculate
and then, binomial theorem can be applied.
It can be written that, 99 100 1
5 5 99 100 1
5 5 5 4 5 3 2 5 2 3 5 4 5 5 C 0 (^) 100 C 1 (^) 100 1 C 2 (^) 100 1 C 3 (^) 100 1 C 4 (^) 100 1 C 51
5 4 3 2 100 5 100 10 100 10 100 5 100 1
Binomial Theorem
8.Binomial Theorem
(^6 6 6 6 5 6 4 6 3 6 2 6 ) x 1 C x 0 C x 1 C x 2 C x 3 C x 4 C x 5 C 6
(^6 6 6 6 5 6 4 6 3 6 2 6 ) x 1 C x 0 C x 1 C x 2 C x 3 C x 4 C x 5 C 6
(^6 6 6 6 6 4 6 2 ) x 1 x 1 2 C x 0 C x 2 C x 4 C 6
6 4 2 2 x 15 x 15 x 1
By putting (^) x 2 we obtain
6 6 6 4 2 2 1 2 1 2 2 15 2 15 2 1
Question 13:
Show that
1 9 8 9
n n
is divisible by 64, whenever n is a positive integer.
Solution 13:
In order to show that
1 9 8 9
n n
(^) is divisible by 64, it has to be prove that,
1 9 8 9 64
n n k
(^) ,
where k is some natural number
By Binomial Theorem,
2 1 0 1 2 ....
m (^) m m m m m a C C a C a C am
For a 8 and m n 1 , we obtain
1 1 1 1 2 1 1 1 8 0 1 8 2 8 .... 1 8
n (^) n n n n n C C C Cn
(^)
1 2 1 1 1 1 9 1 1 8 8 2 3 8 .... 1 8
n n n n n n C C Cn
^
1 1 1 1 1 9 9 8 64 2 3 8 .... 18
n n n n n n C C Cn
^
1 1 1 1 1 9 8 9 64 , where 2 3 8 .... 1 8
n n n n n n k k C C Cn
is a natural number
Thus,
1 9 8 9
n n
(^) is divisible by 64, whenever n is a positive integer.
Question 14:
Prove that
0
3 4
n r n n r r
C
^
Solution 14:
By Binomial Theorem,
0
n n n r r n r r
C a b a b
^ ^
By putting b 3 and a 1 in the above equation, we obtain
Binomial Theorem
8.Binomial Theorem
0
1 3 1 3
n n n^ r^ r^ n r r
C
^ ^
0
3 4
n r n n r r
C
(^)
Hence proved.
Exercise 8.
Question 1:
Find the coefficient of
5
x in
8 x 3
Solution 1:
It is known that 1
th
r term, Tr 1 , in the binomial expansion of
n (^) a b is given by
1
n n r r
Tr C ar b
Assuming that
5
x occurs in the 1
th
r term of the expansion
8 x 3 , we obtain
8 8 1 3
r r Tr Cr x
Comparing the indices of x in
5 x in Tr (^) 1 ,
We obtain r = 3
Thus, the coefficient of
5
x is
8 3 3 3 3
8! 8 7 6 5! 3 3 3 1512 3!5! 3 2.5!
C
.
Question 2:
Find the coefficient of
5 7
a b in
12 a 2 b
Solution 2:
It is known that 1
th
r term, Tr 1 , in the binomial expansion of
n a b is given by
1
n n r r
Tr C ar b
Assuming that
5 7
a b occurs in the 1
th
r term of the expansion
12 a 2 b , we obtain
12 12 12 12 1 2 2
r r r r r Tr Cr a b Cr a b
^ ^ ^
Comparing the indices of a and b in
5 7 (^) a b in Tr (^) 1 ,
We obtain r = 7
Thus, the coefficient of
5 7 a b is
12 7 7 7 7
12! 12 11.10 9 8 7! 2 2 2 792 128 101376 7!5! 5 4.3 2 7!
C
.
Binomial Theorem
8.Binomial Theorem
It is known 1
th
r term, Tr 1 , in the binomial expansion of
n a b is given by 1
n n r r
Tr C ar b
.
Thus, the 13
th term in the expansion of
18 1 9
3
x
x
is
12 18 18 12 13 12 1 12
1 9 3
T T C x x
12 12 12 18!^6 61 1 9 12!6! 3
x
x
(^) (^)
6 12 6 2 6 12 6 12
18 17 16 15 14 13.12! 1 1 .3 9 3 3 12!.6 5 4 3 2 3
x x
(^) (^) ^
Question 7:
Find the middle terms in the expansions of
3 7
3 6
x
Solution 7:
It is known that in the expansion of
n a b , in n is odd, then there are two middle terms,
Namely
1
2
th ^ n^
term and
1 1 2
th ^ n^
term.
Therefore, the middle terms in the expansion
3 7
3 6
(^) x
are
7 1 4 2
th (^) th
and
7 1 1 5 2
th th ^
term
3 3 9 7 7 3^34 4 3 1 (^3 )
7! 3 1 3 6 3!4! 6
x x T T C
(^) (^)
4 9 9 3 3
7 6 5.4! 1 105 3 3 2.4! 2 3 8
x x
3 4 12 7 7 4 4 3 5 4 1 (^4 )
7! 3 1 3 6 4!3! 6
x x T T C
(^) (^)
3 12 12 4 4
7 6 5.4! 3 35
4!.3 2 2 3 48
x x
Thus, the middle terms in the expansion of
7 3
3 6
x
are
(^1059)
8
x and
(^3512)
48
x.
Binomial Theorem
8.Binomial Theorem
Question 8:
Find the middle terms in the expansion of
10
9 3
x y
Solution 8:
It is known that in the expansion of
n (^) a b , in n is even, then the middle term is
1 2
th ^ n
term.
Therefore, the middle term in the expansion of
10
9 3
x y
is
10 1 6 2
th th
10 5 (^5) 10 5 5 5 4 5 1 (^5 )
10! 9 9 3 5!5! 3
x x T T C y y
(^)
5 10 5 5 5 2 10 5
10 9 8 7 6.5! 1 3 9 3 3 5 4 3 2.5! 3
x y
(^) ^
5 5 5 5 5
252 3 x y 6123 x y
Thus, the middle term in the expansion of
10
9 3
x y
is
5 5
61236 x y.
Question 9:
In the expansion of 1
m n a
, prove that coefficients of
m a and
n a are equal.
Solution 9:
It is known that 1
th
r term, Tr 1 , in the binomial expansion of
n a b is given by
1
n n r r
Tr C ar b
Assuming that
m
a occurs in the 1
th
r term of the expansion 1
m n a
, we obtain
1 ^1 ^
m n m n r^ r m n r Tr Cr a C ar
^
Comparing the indices of a in
m a in Tr (^) 1 ,
We obtain r = m
Therefore, the coefficient of
m a is
!! ....... 1 !!!!
m n m
m n m n C m m n m m n
Assuming that
n
a occurs in the 1
th
k term of the expansion 1 ,
m n a
we obtain
1 ^1 ^ ^ ^
m n m n k^ k^ m n k Tk Ck a Ck a
^
Comparing the indices of a in
n
a and in Tk 1 ,
We obtain
k n
Binomial Theorem
8.Binomial Theorem
3
1 5
r
n r
5 r 3 n 3 r 3
3 n 8 r 3 0 ....... 2
Multiplying (1) by 3 and subtracting it from (2), we obtain
4 r 12 0
r 3
Putting the value of r in (1), we obtain n
n 7
Thus, n 7 and r 3
Question 11:
Prove that the coefficient of
n
x in the expansion of
2 1
n x is twice the coefficient of
n x in
the expansion of
2 1 1
n x
(^) .
Solution 11:
It is known that 1
th
r term, Tr 1 , in the binomial expansion of
n a b is given by
1
n n r r
Tr C ar b
^.
Assuming that
n
x occurs in the 1
th
r term of the expansion of
2 1
n x , we obtain
2 2 2 1 1
n n r^ r^ n r Tr Cr x Cr x
^
Comparing the indices of x in
n
x and in Tr 1 , we obtain r n
Therefore, the coefficient of
n
x in the expansion of
2 1
n x is
2 2
2! 2! 2! ....... 1 ! 2!!! (^)!
n n
n n n C n n n n n (^) n
Assuming that
n
x occurs in the 1
th
k term of the expansion of
2 1 1
n x
, we obtain
2 2 1 2 1 1
n n^ k^ k^ n k Tk Ck x Ck x
^
Comparing the indices of x in
n
x and in Tk 1 , we obtain k n
Therefore, the coefficient of
n
x in the expansion of
2 1 1
n x
is
2 1 2 1!^2 1!
! 2 1!! 1!
n n
n n C n n n n n
^
- 2 1! 2! 1 2! ....... 2 2.! 1! 2.!! (^2)!
n n n n
n n n n n (^) n
^ ^ (^)
From (1) and (2) , it is observed that
(^12 2 )
2
n n Cn Cn
Binomial Theorem
8.Binomial Theorem
2 2 1 2
n n Cn Cn
Therefore, the coefficient of
n
x expansion of
2 1
n x is twice the coefficient of
n x in the
expansion of
2 1 1
n x
(^) .
Hence proved.
Question 12:
Find a positive value of m for which the coefficient of
2
x in the expansion 1
m x is 6.
Solution 12:
It is known that 1
th
r term, Tr 1 , in the binomial expansion of
n a b is given by
1
n n r r r r
T C a b
Assuming that
2
x occurs in the 1
th
r term of the expansion of 1
m (^) x , we obtain
1 ^1 ^
m m r^ r^ m r Tr Cr x Cr x
^
Comparing the indices of x in
2 x and in Tr (^) 1 , we obtain r 2
Therefore, the coefficient of
2 x is 2
m
C
It is given that the coefficient of
2
x in the expansion 1
m x is 6.
m
C
! 6 2! 2!
m
m
1 2! 6 2 2!
m m m
m
m m 1 12
2 m m 12 0
2 m 4 m 3 m 12 0
m m 4 3 m 4 0
m 4 m 3 0
m 4 0 or m 3 0
m 4 or m 3
Thus, the positive value of m, for which the coefficient of
2
x in the expansion 1
m x is 6, is
Binomial Theorem
8.Binomial Theorem
From (5), we obtain
5 5 3 3
b b
Thus, a 3, b 5,and n 6.
Question 2:
Find a if the coefficients of
2 x and
3
x in the expansion of
9 3 ax are equal.
Solution 2:
It is known that 1
th
r term, Tr 1 , in the binomial expansion of
n a b is given by
1
n n r r
Tr C ar b
^.
Assuming that
2
x occurs in the 1
th
r term in the expansion of
9 3 ax , we obtain
9 9 9 9 1 3 3
r r r (^) r r Tr Cr ax Cr a x
^
Comparing the indices of x in
2 x and in Tr (^) 1 , we obtain
r 2
Thus, the coefficient of
2 (^) x is
9 9 2^2 7 2 2
9! 3 3 36 3 2!7!
C a a a
Assuming that
3
x occurs in the 1
th
k term in the expansion of
9 3 ax , we obtain
9 9 9 9 1 3 3
k k k (^) k k Tk Ck ax Ck a x
^
Comparing the indices of x in
3
x and in Tk 1 , we obtain k 3
Thus, the coefficient of
3 x is
9 9 3^3 6 3 3
9! 3 3 84 3 3!6!
C a a a
It is given that the coefficient of
2 (^) x and
3 (^) x are the same.
(^6 3 ) 84 3 a 36 3 a
84 a 36 3
36 3 104
84 84
a
9
7
a
Thus, the required value of a is 9/7.
Question 3:
Find the coefficient of
5
x in the product
6 7 1 2 x 1 x using binomial theorem.
Solution 3:
Binomial Theorem
8.Binomial Theorem
Using Binomial Theorem, the expressions,
6 7 1 2 x and 1 x , can be expanded as
6 6 6 6 2 6 3 6 4 6 5 6 6 1 2 x C 0 (^) C 1 (^) 2 x C 2 (^) 2 x C 3 (^) 2 x C 4 (^) 2 x C 5 (^) 2 x C 6 2 x
2 3 4 5 6 1 6 2 x 15 2 x 20 2 x 15 2 x 6 2 x 2 x
2 3 4 5 6 1 12 x 60 x 160 x 240 x 192 x 64 x
7 7 7 7 2 7 3 7 4 7 5 7 6 7 7 1 x C 0 (^) C 1 (^) x C 2 (^) x C 3 (^) x C 4 (^) x C 5 (^) x C 6 (^) x C 7 x
2 3 4 5 6 7 1 7 x 21 x 35 x 35 x 21 x 7 x x
6 7 1 2 x 1 x
2 3 4 5 6 2 3 4 5 6 7 1 12 x 60 x 160 x 240 x 192 x 64 x 1 7 x 21 x 35 x 35 x 21 x 7 x x
The complete multiplication of the two brackets is not required to be carried out. Only those
terms, which involve
5 x , are required.
The terms containing
5 (^) x are
^ ^ ^
5 4 2 3 3 2 4 5 1 21 x 12 x 35 x 60 x 35 x 160 x 21 x 240 x 7 x 192 x 1
5 171 x
Thus, the coefficient of
5 x in the given product is 171.
Question 4:
If a and b are distinct integers, prove that a b is a factor of
n n
a b , whenever n is a positive
integer. [ Hint: write
n n (^) a a b b and expand]
Solution 4:
In order to prove that (^) a b is a factor of
n n a b , it has to be proved that
n n a b k a b , where k is some natural number
It can be written that, a a b b
n n n a a b b a b b
(^1 )
0 1 .... 1
n n^ n n n n n n C a b C a b b Cn a b b C bn
(^) (^)
(^1 ) 1 ... 1
n (^) n n n n n a b C a b b Cn a b b b
(^) (^)
(^1 2 ) 1 .... 1
n n n^ n n n n a b a b a b C a b b Cn b
(^) ^
n n a b k a b
Where,
(^1 2 )
1 .... 1
n (^) n n n n k a b C a b b Cn b
(^)
^
is a natural number
This shows that (^) a b is a factor of
n n
a b , where n is a positive integer.
Binomial Theorem
8.Binomial Theorem
8 6 4 2 2 a 6a 5a 2a 1 8 6 4 2 2a 12a 10a 4a 2
Question 7:
Find an approximation of
5 (^) 0.99 using the first three terms of its expansion.
Solution 7:
5 5 0.99 1 0.
5 5 5 4 5 3 2 C 0 1 C 1 1 0.01 C 2 1 0.01 [Approximately]
2 1 5 0.01 10 0.
Thus, the value of
5 0.99 is approximately 0.951.
Question 8:
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the
expansion of
n
4 4
1 2
3
is 6 :1.
Solution 8:
In the expansion,
n (^) n n n n 1 n n 2 2 n n 1 n n a b C a 0 C a 1 b C a 2 b ... Cn 1 ab C bn
(^)
Fifth term from the beginning
n n 4 4 C a 4 b
Fifth term from the end
n 4 n 4 C a b 4
Therefore, it is evident that in the expansion of
n
4
4
1 2 3
are fifth term from the beginning
is
4 n 4 n 4 (^4 )
1 C 2
3
(^)
and the fifth term from the end is
n 4 4 n 4 n (^4 )
1 C 2
3
n 4 4 n 4 n n 4 n 4 (^4 4 4 ) 4
2 1 1 n! C 2 C 2 ..... 1
3 3 6.4! n^ 4! 2
(^)
4 n 4 4 4 n 4 n n n 4 4 n (^4) n n (^4) n n 4 4 4
1 3 3 6n! 1 C 2 C C .2 ..... 2 3 n^ 4 !4! 3 3 3
Binomial Theorem
8.Binomial Theorem
It is given that the ratio of the fifth term from the beginning to the fifth term from the end is
6 :1. Therefore, from (1) and (2), we obtain
n 4 n 4
n! 6n! 1 2 : 6 : 6.4! n 4! n 4 !4! 3
n 4
n 4
2 6 : 6 : 6 3
n n 4 4 2 3
6 6 6
n 4 6 36 6
n/4 5/ 6 6
n 5
4 2
5 n 4 10 2
Thus, the value of n is 10.
Question 9:
Expand using Binomial Theorem
4 x 2 1 , x 0 2 x
^ ^
Solution 9:
4 x 2 1 2 x
^
4 3 2 2 3 4 n n n n n 0 1 2 3 4
x x 2 x 2 x 2 2 C 1 C 1 C 1 C 1 C 2 2 x 2 x 2 x x
(^) (^) (^) (^) (^) (^) (^) (^)
(^4 3 )
2 3 4
x x 2 x 4 x 8 16 1 4 1 6 1 x 4 1 2 2 x 4 x 2 x x
^ (^) (^) (^) (^) (^) (^) (^) (^)
(^)
4 3
2 3 2 4
x 8 x 24 24 32 16 16 1 1 6 2 x 2 x x x x x
4 3
2 3 4
x 8 x 8 24 32 16 1 1 6 .... 1 2 x 2 x x x x
Again by using Binomial Theorem, we obtain
4 2 3 4 4 4 4 3 4 2 4 3 4 0 1 2 3 4
x x x x x 1 C 1 C 1 C 1 C 1 C 2 2 2 2 2
^ ^ ^ ^ ^