ALGEBRA CHAPTER III BINOMIAL THEOREM
EXERCISE I
1.
Using binomial expansion prove that :
(i)
( x  3 y)3  x3  9 x 2 y  27 xy 2  27 y 3
(ii)
(2 x  3 y)4  16 x 4  96 x3  216 x 2 y 2  216 xy 3  81y 4
5
(iii)
(iv)
1
40 10 1

5
3
 2 x    32 x  80 x  80 x   3  5
x
x x
x

2 4
2
3
4
(1  x  x )  1  4 x  10 x  16 x  19 x  16 x5  16 x6  4 x7  x8
2.
Evaluate (i)
3.
Find
( 3  1)5  ( 3  1)5
(ii)
( 3  2)3  ( 3  2)3 .
12


1

x
(i)
10th term in the expansion of  2x 2 
(ii)
3rd term from the end in the expansion of (2 x  3 y)12
(iii)
1

12 term in  x  
x

15
4.
th
Find the middle term in the following expansions
10
(i)
5.
1

x 
x

10
(ii)
(1  x)
2n
(ii)
1

x in the expansion of  36 x  
6

(iv)
Find the coefficients of :
3
1

in the expansion of  2x 2  
x

(i)
x
(iii)
1

x in the expansion of  x  
x

2
13
7
Find the term independent of x in the expansion of :
12
(i)
7.
8.
8
(iii)
12
6.
a

  2b 
3

1

 2x  
x

77
(ii)
 2 1 
x  
2x 

10
1

 2x  
x

(iii)
 x2 1 
 
Prove that there is no term involving x in the expansion of 
 2 x
9
4
Find the number of terms in the expansion of the following :
(i)
(ii)
(1  2 x  x 2 )20
(2 x  3 y)9
(iii)
[(3x  y)8  (3x  y)8 ]
(v) (2 x  3 y  4 z ) n
(iv) (1  5 2 x)9  (1  5 2 x)9
( x  y )10  ( x  y )10
(vi)
1
6
10.
The first three terms in expansion of ( x  y)n are 1, 56 and 1372 respectively.
Find the values of x & y.
In the expansion of ( x  y)n the coefficients of 4th and 13th terms are equal. Find n.
11.
In the expansion of ( x  y)m n , where m and n are natural numbers, prove that the coefficients
9.
of x m and x n are equal.
12.
13.
14.
The coefficients of (r  1)th , r th and (r  1)th terms in the expansion of (1  x)n are in the ratio
of 1 : 3 : 5 .Find n and r.
The 2nd , 3rd and 4th terms in the expansion of ( x  y)n are 240 , 720 and 1080 respectively .
Find the values of x , y and n.
The coefficients of x r 1 , x r and x r 1 in (1  x) 2 n are in A.P., prove that
n2  n(4r  1)  (4r 2  2)  0 .
15.
Find the value of r if the coefficients of (2r  4)th and (r  2)th terms in the expansion of
(1  x)18 are equal.
16.
Show that the coefficients of the middle term in the expansion of (1  x) 2 n is the sum of the
coefficients of two middle terms in the expansion of (1  x) 2 n 1
b 2  ac 5a
.

c 2  bd 3c
17.
If in the expansion of  x  a  T3 , T4 , T5 , T6 are a, b, c, d respectively show that
18.
Prove that the coefficients of x n in (1  x) 2 n is twice the coefficients of x n in (1  x)2 n1.
19.
Show that coeff. of x 4 in the expansion of 1  2 x   2  x  is 438 .
20.
21.
In the binomial expansion of ( x  1) n , the coefficients of the fifth, sixth and seventh terms are
in arithmetic progression .Find all the values of n for which this can happen.
Show that 20 Cn is greatest when n  10 .
22.
23.
24.
Find a if the coefficients of x 2 and x3 in the expansion of (3  ax)9 are equal.
Find a positive value of m for which the coefficients of x2 in the expansion (1+x)m is 6.
For what value of m the coefficients of the (2m  1)th and (4m  5)th terms, in the expansion of
n
4
5
(1  x)10 , are equal ?
25.
26.
27.
Find a negative value of m if the coefficients of x2 in the expansion of (1+ x)m is 6.
Find two values of m such that the coefficients of x 2 in the expansion of ( -1-x)m is 3.
If P be the sum of odd terms and Q that of even terms in the expansion of ( x  a) 2 n ,
Prove that
(i) ( x 2  a 2 )n  P 2  Q 2 (ii) ( x  a)2 n  ( x  a)2 n  4 PQ
28.
(iii) ( x  a)2 n  ( x  a)2 n  2( P 2  Q2 )
n
n2
2 n 1
Prove that if x lies between
and
the greatest term in the expansion of 1  x 
n2
n
has the greatest coefficient.
29.
If the coefficients of x 7 in  ax 2  1/ bx   is equal to the coefficient of x 7 in
11
ax  1/ bx2  then show that ab  1 .


11
2
1.3.5..... 2n  1 n n
2 .x or Show that term
n!
30.
Show that middle term in the expansion of 1  x  is
31.
1
1.3.5.....2n1 n

independent of x in the expansion of  x   is
2 .
x
n!

Find the value of x for which the sum of the third and the fifth term in the expansion of
2n
n
 x
1
 2 
x1

2

n

 is equal to 135 and the sum of binomial coefficients in the last three terms of


the expansion is equal to 22.
32.

For what values of x is the sixth term in the expansion of  2log2

9 x1  7
2


1
log 2 3x1 1
5
7
 equal

to 84.
33.
If the positive integers r  1, n  2 the coefficients of the  3r  and  r  2  powers of x in the
th
th
expansion of 1  x  are equal, then prove that n  2r  1 .
2n
9
34.
1 
3
Find the term independent of x in the expansion of 1  x  2 x   x 2 
 .
3x 
2
35.
Find out the sum of the coefficients in the expansion of the binomial  5 p  4q  , where n  N .
36.
Determine the value of x in the expression x  xt
3
n

 , if the third term in the expression is
5
10,00,000. Where t  log10 x
m
37.
1/ 5 
 log103x  1/ 2
x  2 log3
  2    is equal to 21
Find the value of x for which the sixth term is   2




and binomial coefficients of second, third and fourth terms are the first, third and fifth terms of
an arithmetic progression. (Take base of log as 10)


6
2  1 is 198 .
38.
Show that integral part of
39.
Show that the integral part of
40.
By using binomial theorem show that :
6n  5n  1 is divisible by 25, n  N . (ii)
(i)
6n  2  7 2 n 1 is divisible by 43.
(iii)
(iv)
2n
5  1 is is divisible by 13 if n is odd (vi)
(v)


5
2  1 is 82.
9n 1  8n  9 is divisible by 64, n  N .
22 n 1  9n 2  3n  2 is divisible by 54.
last three digits of 350 are 249 .
ANSWERS
2.
(i) 152
(ii) 18 3
4.
(i) 252 (ii)
5.
(i) 1760 (ii) - 15 (iii)
(2n)! n
x
(n !) 2
3. (i) 1760x 3 (ii) 15588936x 2 y10 (iii) 1365x 7
(iii) 8064
-8064 6. (i)
1120
(ab) 4
81
495
84 (ii)
(iii) - 8064
256
(iv)
3
(n  1)(n  2)
(vi) 6 9. x  1, y  7
2
n  7, r  3
13. x  2, y  3, n  5 15. 6
n  7,14
22. a  9 / 7 23.
m4
x  2, x  1
26. m  3, 2 31.
m  3
17
35.
1
36. x  10 or 10 5 / 2
54
8.
(i) 41 (ii) 10 (iii) 4 (iv) 5 (v)
10.
19.
24.
n  15
n  15
m 1
12.
20.
25.
32.
x  1, x  2
34.
37.
x  2 or 0
SOLUTIONS (EXERCISE I)
15.
16.
We must have 18 C2r 3  18Cr 3
2r  3  r  3 or 2r  3  r  3  18

r  6 is not possible
r 6


Coefficients of the middle term in the expansion of 1  x 
 2nCn

2n
Tn 1 is the middle term)
Now the two middle terms in the expansion of 1  x 
2 n 1
2 n 1
are Tn and Tn 1 the coefficient of
Cn1 and
Cn which proves the result since n Cr n Cr 1  n1Cr
We have a  nC2 .xn2 y 2 , b n C3 xn3 y3 ,
c n C4 xn4 y 4 , d n C5 xn5 y5
these terms will be
17.
r  6 or r  6
2 n 1
b 2  ac
c 2  bd
n 2 (n  1)2 (n  2) 2 2 n 6 6 n 2 (n  1) 2 (n  2)(n  3) 2 n 6 6

x y 
x y
(3!) 2
2!4!
n 2 (n  1)2 (n  2) 2 (n  3)3 2 n 8 8 n 2 (n  1) 2 (n  2) 2 (n  3)(n  4) 2 n 8 8
x .y 
x y
(4!)2
3!5!
Now
Divide each n containing term by n2 (n  1)2 (n  2) , take xy type terms common and
multiply every term by 5! 4!
This ratio 
x 2 n 6 y 6  (n  2)80  (n  3)60
x 2 n 8 y 8 5(n  2)(n  3)2  4(n  2)(n  3)(n  4) 
20  4n  8  3n  9
20 x 2
n 1
20 x 2
x2

.

.
y 2 (n  2)(n  3) n  1
y 2 (n  2)(n  3)
y 2 (n  2)(n  3) 5(n  3)  4(n  4) 
n(n  1) n  2 2
5.
x y
20 x 2
5a
2

Also RHS 
 2
y (n  2)(n  3)
3c 3 n(n  1)(n  2)(n  3) x n  4 y 4
4!
Hence LHS  RHS

19.
1  2x   2  x 
4
5
  2  x  1  2 x 
5
4
  5 C0 25 ( x)0  5C1 24 ( x)1  5C2 23 ( x)2  5C3 22 ( x)3  5C4 2( x)4  5C5 ( x)5 
1  4C1.2 x  4C2 (2 x)2  4C3 (2 x)3  4C4 (2 x)4 
4
Term of x 4

 5 C0 25 ( x)0 . 4C4 (2x)4    5C1 24 ( x)1. 4C3 (2 x)3 

 
 5 C2 23 ( x)2 . 4C2 (2 x)2 

20.
5
C3 22 ( x)3 . 4C1 (2 x)



 5 C4 2(2 x)4 .1
coefficient of x 4 (leaving x and noting 5 C0  1 etc.)
 32  16  80  32  80  24  40  8  10  438
According to the question n C4 , nC3 , nC6 are in AP
n!
n!
n!


2 nC5  nC4  nC6


(n  5)!5! (n  4)!4! (n  6)!6!
1
1
2
1
1
Cancelling n!,
and
we get


(n  6)!
5(n  5) (n  4)(n  5) 6  5
4!
On multiplying by 30(n  4)(n  5) we get 12(n  4)  30  (n  4)(n  5)

n2  21n  98  0

n  7, n  14
20!
C
20  r 21  (r  1)
21
(r  1)!(19  r )!
We have 20 r 1 



1
20!
Cr
r 1
r 1
r 1
r !(20  r )!
21
Now
will continuously decrease as r runs from 0 to 19 but upto certain stage it will be
r 1
greater than 1 after which it will be less than 1. Let us actually observe
T3
T1
T2
T1  T0
T3  T2  T1  T0
 20 
 9.5  T2  T1  T0 ,
8 
T0
T1
T2
T5
T4
 4.2  T4  T3  T2  T1  T0 ,
 3.2  T5  T4  T3  .....  T0
T3
T4
T6
T7
T7  T6  ......  T0
 2.5  T6  T5  ......  T0 ,
2 
T5
T6
T8
T9
T8  T7  ......  T0 ,
T9  T8  ......  T0
 1.6 
 1.3 
T7
T8
T10
T11
T10  T9  ......  T0 ,
T10  T11
(Note)
 1.1 
 .9 
T9
T10
T12
T12  T11  T10 and so on. It will through out be less than 1)
 .75 
T11
20
21.
T10 is the greatest term 

(Note that 20 C10 is coeff. of 11th term)
24.
27.
n
C2m  nC4m4
 x  a
n

20
C10 is the greatest coefficient
either 2m  4m  4 or 2m  4m  4  10  m  1
 nC0 x n a0  nC1 xn1a n C2 x n2a 2  .... n Cn x0a n
P  nC0 x n a0  nC2 x n2 a2  nC4 xn4 a4  ........
Q n C1 xn1a1  nC3 x n3a3  .......
It is clear that P  Q  ( x  a)n , P  Q  ( x  a)n After which all parts easily follow.
5
28.
The expansion of (1  x)2n1 will contain 2n  2 terms. The two equal greatest coefficients
will be
(1  x)
2 n 1
2 n 1
2 n 1
Cn1 and

2 n 1
C0 
2 n 1
Cn . The expansion can be written as
C1 x  .....  2 n 1 Cn 1 x  ....   2 n 1 Cn x n  2 n 1 Cn 1 x n 1   2 n 1 Cn  2 x n  2
..... 2n1 C2n1
The two terms written in the rectangular block will be greatest terms with equal
coefficients we may write
2 n1
2 n 1
Cn1 xn1 2n1 Cn2 xn2 ;
Cn xn 2n1 Cn1 xn1
(2n  1)!
(2n  1)!
(2n  1)! x
(2n  1)!

x;

(n  1)!n! (n  1)!(n  1)!
(n  1)n! (n  1)!(n  2!)
n2
n
n
n2
Thus x must lie between
and
.
 x;
x
n2
n
n2
n


29.
r
 1 
Tr 1 in first expansion  Cr ax
 11Cr a 11r b r x223r
 
 bx 
r 5
we must have 22  3r  7

coeff. of x7 in the first expansion  11C5 a6b5 (*)

11
 
11 r
2
Again Tr 1 in second expansion  Cr  ax 
We must have 11  3r  7
11
11 r

r 6
 1 
 2 
 bx 
r
r
 Cr a
 1
coeff. of x in the second expansion  C6 a   

 b
11
11
As C5  C6 we easily get by equating (*) and (**) ab  1
7
11
11 r
11
6
5
(**)
Sum of binomial coeffs. in the last three terms  22
31.

n
Cn  nCn1  nCn2  22 
1  nC1  nC2  22 
2  2n  n2  n  44

(n  7)(n  6)  0


Now sum of third and fifth term  135
 
2
1 n 
n2  n  42  0
n6
4
2
 1  6
1 
 C2 2 
 C4 2 x 

  135
x 1
x 1
 2 
 2 
22 x
2x
4
15 x 1  15 2 x  2  135 

2.2x  x  9
2
2
2
which can be easily solved by putting 2x  t .
6
32.
 1  113 r
  x
 b
x

4
log 2 9 x1  7
T6  C5 2
7

 21.2



2


  1 log2 3x1 1 
. 2 5




log2 9x1  7 log2 3x1 1
5
 21.2
 21.22log2
log 2
9 x 1  7
3x 1 1
6
9x1  7


 log2 3x1 1
.2
n(n  1)
 22
2
log
Since T6  84 we get
2
9 x 1  7
3x1 1
 22
9 x 1  7
9 x 1  7
4

2

3x 1  1
3x 1  1
we will easily get x  1,2 by putting 3x 1  t , .

33.
34.
log 2
We must have
2n

C3r 1  2nCr 1
1 
3
In the expansion of  x 2  
3x 
2
 3x 2 
Tr 1  Cr 

 2 
9 r
9
 1 
 
 3x 
r  1 or n  2r  1
But r  1 is given.
9
r
3
 9 Cr  
2
9 r
r
 1  183r
  x
 3
(*)
1
3
Now term independent of x in the expansion 1  x  2 x 3   x 2  
x
2
9
 3x 2 1 
1
3
 
= Term independent of x in  x 2   1. coeff. of x 1 in 
3x 
x
2
 2
9
9
9
 3x 2 1 
2. coeff. of x in 
 
3x 
 2
The first case corresponds to r  6 (By equating 18  3r  0 ). The second one is not
possible (Why?). The third case corresponds to r  7 .
3
3
6
2
35.
 3  1
 3  1
Thus Required answer  9C6       2. 9C7     
 2  3
 2  3
Put p  q  1 to get sum of all the coeffs.
36.
Third term 5 C2 x3 xt
 
2
then x 2t 3  105 
on taking logs we get  2log x  3 log x  5
37.

84
72
17


3
5
8  3 4  3 54
 10 x2t 3
If 10 x2t 3  1000000  106

7
x2log x 3  105
(Base 10)
2 z  3z  5  0 where z  log10 x
2
x  10, x  105 2
z  1, z   5 2 

Let T2 , T3 , T5 be the first, third and fifth term of A.P.
Now co-efficient of T2 m C1  a
say
Coefficient of T3  C2  a  2d
m
Coefficient of T5  mC3  a  4d
Given, 2m C2  mC1   mC3
2(m!)
m!
m!
or
m2  9m  14  0 


(m  2)!2 (m  1)! (m  3)!3!
As 6th term is equal to 21 m  2 is not possible.
Thus m  7 . According to question
7
C5

2log(103)
x
 
7 5
21.2log(103 ).2log(3)
x
x2
5
2( x 2) log3
 21 
m  2 or m  7
  21
5
2log(10 3 )  log 3( x 2)  1  20   log(10  3x ).3( x2)  0
x
7
3x.19  32 x
3x.10  32 x  9  0
1

9
Which can be solved by putting 3x  t .We will easily get x  2,0

38.
Integral part of
40.
(iii)

 
6
2 1 
 6n 2  7.6n  43m
 6n 62  7  43m
(iv)
6
 6n 2  7.49n
6n 2  72 n1

 
2 1 

22n1  9n2  3n  2

6
2 1 1
 0


 6n2  7. 6  43


6
2  1  1

 197
n
(Why?)
 2.4n  9n2  3n  2
 2 1  3  9n2  3n  2
n
n(n  1) 2


 2 1  3n 
3  27m  9n2  3n  2
2!


 54m
(v)
52n  1  25n  1   26  1  1
(vi)
if n is odd (1)n and 1 will cancel
52n  1 is divisible by 13 .

The last three digits of a number = remainder when number is divided by 100
n
Now 350  925
 26m  (1)n  1
 10  1  1000m  last three terms
25
 1000m  25C23102 (1)23 25 C2410(1)24 25 C25 (1)25
25  24
100  250  1
2
 25  24
 1000m  249
100  A multiple of 1000 ).

2

= Remainder when divided by 1000 is 249
Last three digit are 249 .

 1000m 
8