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Binomial Theorem Practice Questions - DPP for JEE

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 Page 2


The term independent of x in above
4. (c) Q  x
3
 and higher powers of x may be neglected   
 
(as x
3
 and higher powers of x can be neglected)
5. (d)  
= [(2 – x) + (2x – 3)]
50
= (x – 1)
50
= (1 – x)
50
= 
50
C
0
 – 
50
C
1
 x ....... – 
50
C
25
 x
25
 + ........
Coefficient of x
25
 is  – 
50
C
25
6. (d) a
0
 + a
1
 + a
2
 + ..... = 2
2n
 and a
0
 + a
2
 + a
4
 + .... = 2
2n –1
a
n
 = 
2n
C
n
 = the greatest coefficient, being the middle coefficient
a
n – 3
 = 
2n
C
n – 3
 = 
2n
C
2n – (n – 3)
 = 
2n
C
n+3
 = a
n +3
7. (c) The number of selection = coefficient of x
8
 in
(1 + x + x
2
 + .... + x
8
) (1 + x + x
2
 + ...... + x
8
). (1 + x)
8
= coefficient of x
8
 in 
Page 3


The term independent of x in above
4. (c) Q  x
3
 and higher powers of x may be neglected   
 
(as x
3
 and higher powers of x can be neglected)
5. (d)  
= [(2 – x) + (2x – 3)]
50
= (x – 1)
50
= (1 – x)
50
= 
50
C
0
 – 
50
C
1
 x ....... – 
50
C
25
 x
25
 + ........
Coefficient of x
25
 is  – 
50
C
25
6. (d) a
0
 + a
1
 + a
2
 + ..... = 2
2n
 and a
0
 + a
2
 + a
4
 + .... = 2
2n –1
a
n
 = 
2n
C
n
 = the greatest coefficient, being the middle coefficient
a
n – 3
 = 
2n
C
n – 3
 = 
2n
C
2n – (n – 3)
 = 
2n
C
n+3
 = a
n +3
7. (c) The number of selection = coefficient of x
8
 in
(1 + x + x
2
 + .... + x
8
) (1 + x + x
2
 + ...... + x
8
). (1 + x)
8
= coefficient of x
8
 in 
= coefficient of x
8
 in  (1 + x)
8
 in (1 + x
8
) (1 – x)
–2
= coefficient of x
8
 in
(
8
C
0
 + 
8
C
1
x + 
8
C
2
x
2
 + ..... + 
8
C
8
x
8
)
× (1 + 2x + 3x
2
 + 4x
3
 + ..... + 9x
8
 +....)
= 9. 
8
C
0
 + 8 · 
8
C
1
 + 7. 
8
C
2
 + .... + 1. 
8
C
8
= C
0
 + 2C
1
 + 3C
2
 + .... + 9C
8
   [C
r
? 
= 
8
C
r
]
Now C
0
x + C
1
x
2
 + .... + C
8
x
9
 = x (1 + x)
8
Differentiating with respect to x, we get
C
0
 + 2C
1
x + 3C
2
x
2
 + .... 9C
8
x
8
 = (1 + x)
8
 + 8x (1 + x)
7
Putting x = 1, we get C
0
 + 2C
1
 + 3C
2
 + ..... + 9C
8
= 2
8
 + 8.2
7
. = 2
7
 (2 + 8) = 10.2
7
.
8. (b)
 given expression
= 
we know that,
Which is a polynomial of degree 6.
9. (a) To find
30
C
0
? 30
C
10
 – 
30
C
1
? 30 
C
11
 + 
30 
C
2
? 30
C
12
 – .... + 
30
C
20
? 30
C
30
We know that
(1 + x)
30
 = 
30
C
0
 + 
30
C
1
x + 
30
C
2
x
2
+ .... + 
30
C
20
x
20
 + ....
30
C
30
x
30
....(1)
(x – 1)
30
 = 
30
C
0
x
30
 – 
30
C
1
x
29
 +....+ 
30
C
10
x
20
–
 30
C
11
x
19
 + 
30
C
12
x
18
 +.... 
30
C
30
x
0
....(2)
Multiplying eq
n
 (1) and (2) and equating the coefficients of x
20
 on both
sides, we get
Page 4


The term independent of x in above
4. (c) Q  x
3
 and higher powers of x may be neglected   
 
(as x
3
 and higher powers of x can be neglected)
5. (d)  
= [(2 – x) + (2x – 3)]
50
= (x – 1)
50
= (1 – x)
50
= 
50
C
0
 – 
50
C
1
 x ....... – 
50
C
25
 x
25
 + ........
Coefficient of x
25
 is  – 
50
C
25
6. (d) a
0
 + a
1
 + a
2
 + ..... = 2
2n
 and a
0
 + a
2
 + a
4
 + .... = 2
2n –1
a
n
 = 
2n
C
n
 = the greatest coefficient, being the middle coefficient
a
n – 3
 = 
2n
C
n – 3
 = 
2n
C
2n – (n – 3)
 = 
2n
C
n+3
 = a
n +3
7. (c) The number of selection = coefficient of x
8
 in
(1 + x + x
2
 + .... + x
8
) (1 + x + x
2
 + ...... + x
8
). (1 + x)
8
= coefficient of x
8
 in 
= coefficient of x
8
 in  (1 + x)
8
 in (1 + x
8
) (1 – x)
–2
= coefficient of x
8
 in
(
8
C
0
 + 
8
C
1
x + 
8
C
2
x
2
 + ..... + 
8
C
8
x
8
)
× (1 + 2x + 3x
2
 + 4x
3
 + ..... + 9x
8
 +....)
= 9. 
8
C
0
 + 8 · 
8
C
1
 + 7. 
8
C
2
 + .... + 1. 
8
C
8
= C
0
 + 2C
1
 + 3C
2
 + .... + 9C
8
   [C
r
? 
= 
8
C
r
]
Now C
0
x + C
1
x
2
 + .... + C
8
x
9
 = x (1 + x)
8
Differentiating with respect to x, we get
C
0
 + 2C
1
x + 3C
2
x
2
 + .... 9C
8
x
8
 = (1 + x)
8
 + 8x (1 + x)
7
Putting x = 1, we get C
0
 + 2C
1
 + 3C
2
 + ..... + 9C
8
= 2
8
 + 8.2
7
. = 2
7
 (2 + 8) = 10.2
7
.
8. (b)
 given expression
= 
we know that,
Which is a polynomial of degree 6.
9. (a) To find
30
C
0
? 30
C
10
 – 
30
C
1
? 30 
C
11
 + 
30 
C
2
? 30
C
12
 – .... + 
30
C
20
? 30
C
30
We know that
(1 + x)
30
 = 
30
C
0
 + 
30
C
1
x + 
30
C
2
x
2
+ .... + 
30
C
20
x
20
 + ....
30
C
30
x
30
....(1)
(x – 1)
30
 = 
30
C
0
x
30
 – 
30
C
1
x
29
 +....+ 
30
C
10
x
20
–
 30
C
11
x
19
 + 
30
C
12
x
18
 +.... 
30
C
30
x
0
....(2)
Multiplying eq
n
 (1) and (2) and equating the coefficients of x
20
 on both
sides, we get
30
C
10
 = 
30
C
0
? 30 
C
10
 – 
30 
C
1
? 30
C
11 
+ 
30
C
2
? 30
C
12
– ....+ 
30
C
20  
? 30
C
30
? Req. value is 
30
C
10
10. (c) Let the consecutive coefficient of
(1 + x)
n
 are  
n
C
r–1
, 
n
C
r
, 
n
C
r + 1
From the given condition,
n
C
r–1 
:  
n
C
r
 : 
n
C
r + 1
 = 6 : 33 : 110
Now 
n
C
r – 1
: 
n
C
r  
=
 
6 : 33
 
  11r = 2n – 2r + 2
 2n – 13r + 2 = 0 ....(i)
and 
n
C
r 
:  
n
C
r + 1
= 33 : 110
 
   3n – 13r – 10 = 0 ...(ii)
Solving (i) & (ii), we get n = 12
11. (d) in the expansion
= 
For the coefficient of x
7
, we have
22 – 3r = 7 r = 5
  Coefficient of x
7
 = ...(i)
Again T
r+1
 in the expansion
= 
For the coefficient of x
–7
, we have
Page 5


The term independent of x in above
4. (c) Q  x
3
 and higher powers of x may be neglected   
 
(as x
3
 and higher powers of x can be neglected)
5. (d)  
= [(2 – x) + (2x – 3)]
50
= (x – 1)
50
= (1 – x)
50
= 
50
C
0
 – 
50
C
1
 x ....... – 
50
C
25
 x
25
 + ........
Coefficient of x
25
 is  – 
50
C
25
6. (d) a
0
 + a
1
 + a
2
 + ..... = 2
2n
 and a
0
 + a
2
 + a
4
 + .... = 2
2n –1
a
n
 = 
2n
C
n
 = the greatest coefficient, being the middle coefficient
a
n – 3
 = 
2n
C
n – 3
 = 
2n
C
2n – (n – 3)
 = 
2n
C
n+3
 = a
n +3
7. (c) The number of selection = coefficient of x
8
 in
(1 + x + x
2
 + .... + x
8
) (1 + x + x
2
 + ...... + x
8
). (1 + x)
8
= coefficient of x
8
 in 
= coefficient of x
8
 in  (1 + x)
8
 in (1 + x
8
) (1 – x)
–2
= coefficient of x
8
 in
(
8
C
0
 + 
8
C
1
x + 
8
C
2
x
2
 + ..... + 
8
C
8
x
8
)
× (1 + 2x + 3x
2
 + 4x
3
 + ..... + 9x
8
 +....)
= 9. 
8
C
0
 + 8 · 
8
C
1
 + 7. 
8
C
2
 + .... + 1. 
8
C
8
= C
0
 + 2C
1
 + 3C
2
 + .... + 9C
8
   [C
r
? 
= 
8
C
r
]
Now C
0
x + C
1
x
2
 + .... + C
8
x
9
 = x (1 + x)
8
Differentiating with respect to x, we get
C
0
 + 2C
1
x + 3C
2
x
2
 + .... 9C
8
x
8
 = (1 + x)
8
 + 8x (1 + x)
7
Putting x = 1, we get C
0
 + 2C
1
 + 3C
2
 + ..... + 9C
8
= 2
8
 + 8.2
7
. = 2
7
 (2 + 8) = 10.2
7
.
8. (b)
 given expression
= 
we know that,
Which is a polynomial of degree 6.
9. (a) To find
30
C
0
? 30
C
10
 – 
30
C
1
? 30 
C
11
 + 
30 
C
2
? 30
C
12
 – .... + 
30
C
20
? 30
C
30
We know that
(1 + x)
30
 = 
30
C
0
 + 
30
C
1
x + 
30
C
2
x
2
+ .... + 
30
C
20
x
20
 + ....
30
C
30
x
30
....(1)
(x – 1)
30
 = 
30
C
0
x
30
 – 
30
C
1
x
29
 +....+ 
30
C
10
x
20
–
 30
C
11
x
19
 + 
30
C
12
x
18
 +.... 
30
C
30
x
0
....(2)
Multiplying eq
n
 (1) and (2) and equating the coefficients of x
20
 on both
sides, we get
30
C
10
 = 
30
C
0
? 30 
C
10
 – 
30 
C
1
? 30
C
11 
+ 
30
C
2
? 30
C
12
– ....+ 
30
C
20  
? 30
C
30
? Req. value is 
30
C
10
10. (c) Let the consecutive coefficient of
(1 + x)
n
 are  
n
C
r–1
, 
n
C
r
, 
n
C
r + 1
From the given condition,
n
C
r–1 
:  
n
C
r
 : 
n
C
r + 1
 = 6 : 33 : 110
Now 
n
C
r – 1
: 
n
C
r  
=
 
6 : 33
 
  11r = 2n – 2r + 2
 2n – 13r + 2 = 0 ....(i)
and 
n
C
r 
:  
n
C
r + 1
= 33 : 110
 
   3n – 13r – 10 = 0 ...(ii)
Solving (i) & (ii), we get n = 12
11. (d) in the expansion
= 
For the coefficient of x
7
, we have
22 – 3r = 7 r = 5
  Coefficient of x
7
 = ...(i)
Again T
r+1
 in the expansion
= 
For the coefficient of x
–7
, we have
11 – 3r = – 7 3r = 18    r = 6
 Coefficient of 
 Coefficient of x
7
 = Coefficient of x
–7
 ab = 1.
12. (d) We have
 + 
 + .....+ 
= 
= 
General Term 
= 
13. (a) (1 + x)
4n
 = 
4n
C
0
 + 
4n
C
1
 x + 
4n
C
2 
x
2
 +
4n
C
3 
x
3
+ 
4n
C
4 
x
4
 + ..... + 
4n
C
un  
x
4n
Put x = 1 and x = – 1, then adding.
2
4n–1
 = 
4n
C
0
 + 
4n
C
2
 + 
4n
C
4
 + ..... + 
4n
C
4n
..... (i)
Now put, x = i
(1 + i)
4n
 = 
4n
C
0
 + 
4n
C
1
i – 
4n
C
2
 – 
4n
C
3
i + 
4n
C
4
 + ........ + 
4n
C
4n
Compare real and imaginary part, we get
(–1)
n
 (2)
2n
 = 
4n
C
0
 – 
4n
C
2
 + 
4n
C
4
 – 
4n
C
6
 + .... + 
4n
C
4n 
... (ii)
Adding (i) and (ii), we get
?
4n
C
0
 + 
4n
C
4
 + .... + 
4n
C
4n
 = (–1)
n
 (2)
2n–1
 + 2
4n–2

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