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BINOMIAL THEOREM
PART 2 OF 2
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Some questions (Assertion–Reason type) are given below. Each question contains Statement – 1 (Assertion)
and Statement – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is
correct. So select the correct choice :Choices are :
(A)Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1.
(B)Statement – 1 is True, Statmnt – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.
(C)
Statement – 1 is True, Statement – 2 is False.
(D)
Statement – 1 is False, Statement – 2 is True.
BINOMIAL THEOREM
373.
Statement-1: The binomial theorem provides an expansion for the expression (a + b)n. where a, b, n  R.
Statement-2: All coefficients in a binomial expansion may be obtained by Pascal’s triangle.
374.
Statement-1: If n is an odd prime then integral part of ( 5  2) n  2 n 1 ([x] is divisible by 20 n.
Statement-2: If n is prime then nC1, nC2, nC3, ….. nCn – 1 must be divisible by n.
375.
376.
377.
378.
Statement–1 : 260 when divided by 7 leaves the reminder 1.
Statement–2 : (1 + x)n = 1 + n1x, where n, n1  N.
Statement–1 : 21 C 0  21 C1  ...  21 C10  220
Statement–2 : 2n 1 C 0  2n 1 C1  ...2n 1 C 2n 1  2 2n 1 and nCr = nC n – r
Let n be a positive integers and k be a whole number, k ≤ 2n.
Statement–1 : The maximum value of 2nC k is 2nC n.
2n
C
Statement–2 : 2 n k 1  1 , for k = 0, 1, 2, . . . , n – 1.
Ck
Let n be a positive integer. Statement–1
:
32n  2  8n  9 is divisible by 64.
n 1
Statement–2 : 32n  2  8n  9  1  8   8n  9 and in the binomial expansion of (1+8)n+1,
sum of first two terms is 8n + 9 and after that each term is a multiple of 82.
379.
380.
381.
Statement–1 : If n is an odd prime, then integral part of

52

n
is divisible by 20n.
Statement–2 : If n is prime, then nc1, nc2, nc3 . . . nc n – 1 must be divisible by n.
Statement–1 : The coefficient of x203 in the expression (x – 1)(x2 – 2) (x2 – 3) . . . (x 20 – 20)
must be 13.
Statement–2 : The coefficient of x8 in the expression (2 + x)2 (3 + x)3 (4 + x)4 is equal to 30.
2n!
Statement–1 : C02 + C12 + C 22 + C32 + ... + Cn2 =
Statement–2 : nC0 – nC1 + nC 2 .... + (–1)n nCn = 0
(n!) 2
382.
Statement–1 : Some of coefficient (x – 2y + 4z)n is 3 n
Statement–2 : Some of coefficient of (c0x0 + c1x1 + c2x 2 + ..... + cnxn)n is 2n
383.
Statement-1: The greatest coefficient in the expansion of (a1 + a2 + a3 + a4)17 is
17!
(3!)3 4!
Statement-2: The number of distinct terms in (1 + x + x2 + x3 + x4 + x5)100 is 501.
21 of 25
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BINOMIAL THEOREM
PART 2 OF 2
384.
385.
386.
387.
Statement-1:
Statement-2:
Statement-1:
Statement-2:
Statement-1:
Statement-2:
are equal.
The co-efficient of x5 in the expansion of (1 + x2)5 (1 + x)4 is 120
The sum of the coefficients in the expansion of (1 + 2x – 3y + 5z)3 is 125.
The number of distinct terms in (1 + x + x2 + x3 + x4)1000 is 4001
The number of distinct terms in the expansion (a1 + a2 + ... + am)n is n+m-1Cm-1
In the expansion of (1 + x)30, greatest binomial coefficient is 30C15
In the expansion of (1 + x)30, the binomial coefficients of equidistant terms from end & beginning
Statement-1: Integral part of


3 1
2n 1
is even where nI.
Statement-2: Integral part of any integral power of the expression of the form of p +
20
388.
Statement-1 :
389.
Statement-1:
Statement-2:
Statement-1:
Statement-2:

r
C 4 = 21C 4 Statement-2: 1 + x + x2 + x3 + ... + xn-1 =
r 4
390.
391.
392.
393.
q is even.
1 xn
= sum of n terms of GP.
1 x
Last two digits of the number (13)41 are 31.
When a number in divided by 1000, the remainder gives the last three digits.
n
C0 + nC1 + nC2 + ….. + nCn = 2 n where n  N.
The all possible selections of n distinct objects are 2 n.
n
Statement-1 : The integral part of (5  2 6) is odd, where n  N.
Statement-2 : (x + a)n  (x  a)n = 2[nC0xn + nC2xx  2 a2 + nC4 + xn  4 a4 + …..]
Statement-1: If n is even than 2nC1 + 2nC3 + 2nC5 + ... + 2nCn-1 = 22n-1
Statement-2: 2nC1 + 2nC3 + 2nC5 + ... + 2nC2n-1 = 22n-1
Statement-1 : Any positive integral power of


N  N  1 for some natural
2  1 can be expressed as
number N > 1.
Statement-2 : Any positive integral power of
integers.
2  1 can be expressed as A + B
2 where A and B are
m
394.
395.
396.
1
4m!


Statement-1 : The term independent of x in the expansion of  x   3  is
.
x
(2m!) 2


Statement-2: The Coefficient of xb in the expansion of (1 + x)n is nCb.
Statement-1: The coefficient of x8 in the expansion of (1 + 3x + 3x2 + x3)17 is 51C2.
Statement-2 : Coefficient of xr in the expansion of (1 + x)n is nCr.
Statement-1: If (1 + x)n = c0+c1x + c2x2 + … + cnxn then
c0  2.c1 + 3.c2 ….. + (1)n (n + 1)cn = 0
Statement-2: Coefficients of equidistant terms in the expansion of (x + a)n where n  N are equal.
n
397.
Statement-1:
k
n
2
Cn   n
2n 1
C n 1
k 1
398.
Statement-2: If 22003 is divided by 15 then remainder is 8.
Statement-1: The co-efficient of (1 + x2)5 (1 + x)4 is 120.
10
Statement-2: The integral part of ( 5  2)
is odd.
ANSWER
373. D 374. A
375. A 376. A 377. A
378. A 379. A 380. C 381. B 382. C 383. D
384. D 385. B 386. B 387. C 388. D 389. D 390. A
391. B. 392. D 393. A 394. D 395. D
396. B
397. B
398. D
399. A
QUE. FROM COMPT. EXAMS.
1.
2.
3.
The value of ( 2  1)6  ( 2  1)6 will be
[RPET 1997]
(a) – 198
(b) 198
(c)
If (1  ax)n  1  8 x  24 x 2  ...., then the value of a and n is
(a) 2, 4
(b) 2, 3
(c)
5
2 5
The coefficient of x in the expansion of (1  x ) (1  x) 4 is
22 of 25
99
(d)
– 99
[IIT 1983; Pb. CET 1994, 99]
3, 6
(d)
1, 2
[EAMCET 1996; UPSEAT 2001; Pb. CET 2002]
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BINOMIAL THEOREM
PART 2 OF 2
(a) 30
(b) 60
1/ 2
4.
If
(1  3 x)
(c)
40
(d)
None of these
5/3
 (1  x)
4x
is approximately equal to a  bx for small values of x, then (a, b) =
 35 

 24 

(a) 1,
(b)  1,

35 

24 
(c)
 35 
 2,
 (d)
 12 
35 

 2,

12 

(x )
5.
The value of x in the expression [x  x log10 ]5 , if the third term in the expansion is 10,00,000
(a) 10
(b) 11
(c)
12
(d)
None of these
6.
If the coefficient of the middle term in the expansion of (1  x)2n 2 is p and the coefficients of middle terms in the
expansion of (1  x)2n1 are q and r, then
(a) p  q  r
(b) p  r  q
(c)
(d)
pqr
pqr 0
7.
In the polynomial ( x  1)(x  2)(x  3)......... ....( x  100), the coefficient of x 99 is
(a) 5050
(b) – 5050
(c)
100
(d)
8.
The coefficient of x 100 in the expansion of
[Roorkee 1992]
[AMU 2002]
99
200
 (1  x) j
is
j 0
[UPSEAT 2004]
 200 

(a) 
100 
9.
 201 

(b) 
102 

If the coefficient of x 7 in  ax 2 

1 

bx 
(c)
 200 


101 
11
 201 


100 
(d)

11
is equal to the coefficient of x 7 in  ax 

1 
 , then ab =
bx 2 
[MP PET 1999; AMU 2001; Pb. CET 2002; AIEEE 2005]
(a) 1
10.
11.
12.
(b) 1/2
(c)
2
(d)
3
5

If the coefficient of x in the expansion of  x 2 

k
 is 270, then k =
x
[EAMCET 2002]
(a) 1
(b) 2
(c)
3
(d)
4
n
The coefficients of three successive terms in the expansion of (1  x) are 165, 330 and 462 respectively, then the
value of n will be
[UPSEAT 1999]
(a) 11
(b) 10
(c)
12
(d)
8
th
th
18
If the coefficient of (2r  4) and (r  2) terms in the expansion of (1  x) are equal, then r=
[MP PET 1997; Pb. CET 2001]
(a) 12
13.
(b) 10
(c)
8
(d)
6
(c)
1.3.5....(2n  1) n
x
n!
The middle term in the expansion of (1  x) 2n is
[Pb. CET 1998]
1.3.5....(5n  1) n
(a)
x
n!
14.
2.4.6.... 2n 2n1
(b)
x
n!
1.3.5....(2n  1) n n
(d)
2 x
n!
 30   30   30   30 
       
 0   10   1  11 
The value of 
 30   30 
 30   30 
    .......     
 2  12 
 20   30 
+ 
[IIT Screening 2005]
(a)
15.
60
C20
(b)
30
C10
(c)
60
(c)
10 th
(c)
462x and
C30
(d)
40
C30
Middle term in the expansion of (1  3 x  3 x 2  x 3 )6 is
[MP PET 1997]
(a) 4 th
16.
(b) 3rd
1
Two middle terms in the expansion of  x  

(a) 231x and
231
x
x
(b) 462x and
(d)
None of these
11
are
462
x
23 of 25
462
(d)
x
None of these
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BINOMIAL THEOREM
PART 2 OF 2
17.
18.
The term independent of y in the expansion of (y 1 / 6  y1 / 3 )9 is
(a) 84
(b) 8.4
(c)
0.84
3
2
– 84
9
The coefficient of the term independent of x in the expansion of (1  x  2x 3 ) x 2 
(a)
19.
[BIT Ranchi 1980]
(d)
1
3
(b)
19
54
 x
3
The term independent of x in 
 2
3
x


(c)
17
54
(c)
4
3
(d)
1 
 is [DCE 1994]
3x 
1
4
10
is
[EAMCET 1984; RPET 2000]
(a)
20.
2
3
5
(b)
3
2
x

(d)
None of these
18
The term independent of x in  x   is

[EAMCET 1990]
(a)
21.
18
C6 2 6
(b)
(b) 51st
C1
C
C
C
 2 2  3 3  ....  15 15 
C0
C1
C2
C14
(a) 100
23.
(b) 120
18
(c)
C18 218
(d)
1
is
5
(c)
[IIT Screening 1993]
7th
(d)
6th
120
(d)
None of these
3n
(d)
None of these
[IIT 1962]
(c)
(b) 0
(c)
n
If Cr stands for Cr , the sum of the given series
2(n / 2)!(n / 2)! 2
[C0  2C12  3C 22  .....  (1)n (n  1)Cn2 ] , Where n is an even positive integer, is
n!
(b) (1)n / 2 (n  1)
(a) 0
25.
1
(x  a) 2n  (x  a) 2n
4
(c) 4 AB  (x  a)2n  (x  a)2n
(1)n / 2 (n  2)
[RPET 1987; UPSEAT 2004]
(b)
2 AB  (x  a) 2n  ( x  a) 2n
(d)
None of these
(b) ( x 2  a 2 )n
(c)
(b) 1
(c)
( x  a)2n (d)
( x  a) 2n
1
[IIT 1982]
(d)
2 2163
If the sum of the coefficients in the expansion of (1  3 x  10 x 2 )n is a and if the sum of the coefficients in the
expansion of (1  x 2 )n is b, then
(a) a  3b
29.
(d)
The sum of the coefficients in the expansion of (1  x  3 x 2 )2163 will be
(a) 0
28.
(1)n (n  2)
[IIT 1986]
In the expansion of ( x  a)n , the sum of odd terms is P and sum of even terms is Q, then the value of (P 2  Q 2 ) will
be
[RPET 1997; Pb. CET 1998]
(a) ( x 2  a 2 )n
27.
(c)
Sum of odd terms is A and sum of even terms is B in the expansion ( x  a)n , then
(a) AB 
26.
None of these
 n
 n
 n
 n
   2    2 2    .....  2 n   is equal to [AMU 2000]
0 
1
 2
 n
(a) 2n
24.
C6 212
The largest term in the expansion of (3  2x)50 where x 
(a) 5th
22.
18
(b) a  b 3
[UPSEAT 2001]
(c)
b  a3
(d)
None of these
The sum of the coefficients in the expansion of ( x  y)n is 4096. The greatest coefficient in the expansion is
[Kurukshetra CEE 1998; AIEEE 2002]
24 of 25
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BINOMIAL THEOREM
PART 2 OF 2
(a) 1024
30.
(b) 924
(c)
824
(d)
724
If the sum of the coefficients in the expansion of (x 2  2 x  1)35 is equal to the sum of the coefficients
in the expansion of ( x  y)35 , then  =
(a) 0
(c) May be any real number
31.
For every natural number n, 3
2n  2
(b)
1
(d)
No such value exist
 8n  9 is divisible by
[IIT 1977]
(a) 16
32.
(b) 128
(c)
256
(d)
None of these
The least remainder when 17 30 is divided by 5 is
[Karnataka CET 2003]
(a) 1
33.
(b) 2
3
(d)
4
n
The value of the natural numbers n such that the inequality 2  2n  1 is valid is
(a) For n  3
34.
(c)
(b) For n < 3
(c)
[MNR 1994]
For mn (d)
For any n
Let P(n) be a statement and let P(n)  p(n + 1) for all natural numbers n, then P(n) is true
(a) For all n
(b)
For all n > 1
(c) For all n > m, m being a fixed positive integer
(d) Nothing can be said
35.
(1  x)n  nx  1 is divisible by (where n  N )
(b) x 2
(a) 2 x
2x 3
(c)
(d)
All of these
ANSWER KEY
1
b
2
a
3
b
4
b
5
a
6
c
7
b
8
a
9
a
10
c
11
a
12
d
13
d
14
b
15
c
16
c
17
d
18
c
19
b
20
a
21
c
22
b
23
c
24
d
25
c
26
b
27
c
28
b
29
b
30
b
31
a
32
d
33
a
34
d
35
b
for 39 Yrs. Que. of IIT-JEE
&
15 Yrs. Que. of AIEEE
we have distributed already a book
25 of 25