unit 6 binomial theorem
TRANSCRIPT
7/29/2019 Unit 6 Binomial Theorem
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SYLLABUS
Mathematical induction and its applications, Binomial theorem for a positive integral index,properties of binomial coefficients,
1. MATHEMATICAL INDUCTION AND ITS APPLICATIONS
It is often used to prove a statement depending upon a natural number n.Type I: If P(n) is a statement depending upon n, then to prove it by induction, we proceedas follows:
(i) Verify the validity of P(n) for n = 1.(ii) Assume that P(n) is true for some positive integer m and then using it
establish the validity of P(n) for n = m + 1.
Then, P(n) is true for each n N.
Illustration 1: Prove that if sin 0, thenn 1
n
n 1
sin2cos cos2 cos4 ...... cos2
2 sin
+
+
aa × a × a a =
a, holds for each n n.
Solution: If P(n) denotes the given statement, then for n = 1, P(1):
sin4cos cos2
4sin
aa × a =
a, which is true
becausesin4 2sin2 cos2
4sin 4sin
a a a=
a a
2 2sin cos cos2
4sin
´ a a a=
a
= cos cos 2.Suppose that P(n) is true for some positive integer m,
i.e.m 1
m
m 1
sin2cos cos2 ...... cos2
2 sin
+
+
aa × a a =
a
Using (1), we shall prove P(n) is true for n = m + 1
i.e.m 2
m m 1
m 2
sin2cos cos2 ...... cos2 cos2
2 sin
+
+
+
aa × a a × a =
a
L.H.S.m 1
m 1
m 1
sin2 sincos2
2 sin
+
+
+
a= × a
a
m 1 m 1
m 12sin2 cos2
2 2 sin
+ +
+
a=× a
m 1
m 2
sin(2 2 )
2 sin
+
+
× a= ×
aR.H.S.
Hence, P(n) is true for each n.
Type II: If P(n) is a statement depending upon n but beginning with some positive integer k,then to prove P(n), we proceed as follows:
(i) Verify the validity of P(n) for n = k.
(ii) Assume that the statement is true for n = m k. Then, using it establish the
validity of P(n) for n = m + 1.Then, P(n) is true for each n k
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Illustration 2: Prove the inequality:n
2
4 (2n) !
n 1 (n !)<
+
, for n 2.
Solution : Let P(n) :n
2
4 (2n) !
n 1 (n !)<
+.
For n = 2, P(2):2
24 4 !
2 1 (2)<
+or 16 24
3 4£
which is true.
Suppose that P(m) is true for n = m 2
i.e.m
2
4 (2m) !
m 1 (m !)<
+. . . (1)
Using (1), we shall prove P(m + 1)
i.e.m 1
2
4 (2(m 1))!
m 2 ((m 1)!)
++
<+ +
L.H.S.m 1 m
24 4 4(m 1) (2m) ! 4(m 1)
m 2 m 1 m 2 (m !) m 2
+
+ += = × < ×+ + + +
[Using (1)]2
2 2
(2m)! (2m 1) (2m 2) 4(m 1) (m 1)
(2m 1) (2m 2) (m!) (m 1) (m 2)
+ + + +=
+ + + +
2
2 2
(2(m 1))! 2(m 1) (2(m 1))!
((k 1)!) (2m 1) (m 2) ((m 1)!)
+ + += × <
+ + + +
2 2
2
2(m 1) 2m 4m 2
(2m 1)(m 2) 2m 5m 2
+ + +=
+ + + +
Hence, P(n) is true for n 2
Note: 1. Product of r consecutive integers is divisible by r !.
2. For x y, xn – yn is divisible by(i) x + y if n is even(ii) x – y if n is even or odd.
2. BINOMIAL EXPRESSION An algebraic expression containing two terms is called a binomial expression.
For example, (a + b), (2x – 3y),2
1 3 2 1x , x ,
y x x x
etc. are binomial expressions.
BINOMIAL THEOREM FOR POSITIVE INDEXSuch formula by which any power of a binomial expression can be expanded in the form of a series is known as Binomial Theorem. For a positive integer n , the expansion is given by
(a+x)n = nC0an + nC1a
n –1 x + nC2 an-2 x2 + . . . + nCr an –r xr + . . . + nCnx
n =
n
0r
rrn
r
nxaC .
where nC0 , nC1 , nC2 , . . . , nCn are called Binomial co-efficients. Similarly
(a – x)n = nC0an – nC1a
n –1 x + nC2 an-2 x2 – . . . + ( –1)r nCr an –r xr + . . . +( –1)n nCnx
n
i.e. (a – x)n =
n
0r
rrnr
nrxaC1
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Replacing a = 1, we get
(1 + x)n = nC0 +nC1x+nC2x2 + . . . + nCr
xr + . . . + nCnxn
and (1 – x)n = nC0 –nC1x+nC2x
2 – . . . + ( –1)r nCr xr + . . . +( –1)n nCnx
n
Observations:
There are (n+1) terms in the expansion of (a +x)n
. Sum of powers of x and a in each term in the expansion of (a +x)n is constant and
equal to n.
The general term in the expansion of ( a+x)n is (r+1)th term given as Tr+1 = nCr an-r xr
The pth term from the end = ( n –p + 2)th term from the beginning .
Coefficient of xr in expansion of (a + x)n is nCr an - r xr .
nCx = nCy x = y or x + y = n.
In the expansion of (a + x)n and (a –x)n, xr occurs in (r + 1)th term.
Illustration 3: If the coefficients of the second, third and fourth terms in theexpansion of (1 + x)n are in A.P., show that n = 7.
Solution: According to the question nC1 nC2
nC3 are in A.P.2n(n 1) n(n 1)(n 2)
n2 6
n2 – 9n + 14 = 0 (n – 2)(n – 7) = 0 n = 2 or 7
Since the symbol nC3 demands that n should be 3
n cannot be 2, n = 7 only.
Illustration 4: Find the(i) last digit (ii) last two digit (iii) last three digit of 17256.
Solution: 17256 = 289128 = (290 –1)128 = 128C0(290)128 –128C1(290)127 + ………..+ 128C126(290)2 –128C127(290)+1= 1000m + 128C2(290)2 –128C1(290) + 1
= 1000m +2
127128(290)2 –
1
290128+ 1 = 1000m + 683527680 + 1
Hence the last digit is 1. Last two digits is 81. Last three digit is 681.
Illustration 5: If the binomial coefficients of (2r + 4)th, (r –2)th term in theexpansion of (a + bx)18 are equal find r.
Solution: This is possible only wheneither 2r + 3 = r –3 …….(1) or 2r + 3 + r –3 = 18 ……..(2) from (1) r = –6 not possible but from (2) r = 6Hence r = 6 is the only solution.
Illustration 6: Find the coefficient of (i) x7 in
11
2 1ax
bx
, (ii) and x –7 in
11
2
1ax
bx. Find the relation between a and b if these coefficients
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are equal.
Solution : The general term in
11 r
2 11 2 11 r
r
1 1ax C (ax )
bx bx
=11 r
11 22 3r
r r
aC x
b
If in this term power of x is 7, then 22 – 3r = 7 r = 5
coefficient of x7 =6
11
5 5
aC
b …(1)
The general term in
11 r
r 11 11 r
r 2 2
1 1ax ( 1) C (ax)
bx bx
=11 r
r 11 11 3r
r r
a( 1) C x
b
If in this term power of x is –7, then 11 – 3r = –7 r = 6
coefficient of x –7 = ( –1)6 11 6 5
11 11
6 56 6
a aC C
b b
If these two coefficient are equal, then6 5
11 11
5 55 6
a aC C
b b
6 6 5 5 5 5a b a b a b (ab 1) 0 ab 1(a 0, b 0)
MIDDLE TERMThere are two cases
(a) When n is evenClearly in this case we have only one middle term namely Tn/2 + 1. Thus middle term in theexpansion of (a + x)n will be nCn/2 an/2xn/2 term.
(b) When n is odd
Clearly in this case we have two middle terms namely2
3
2
1 nnT and T . That means the
middle terms in the expansion of (a +x)n are 2
1n
2
1n
2
1nn x.a.C
and 2
1n
2
1n
2
1nn x.a.C
.
Illustration 7: Find the middle term in the expansion of 9
3
6
xx3
.
Solution: There will be two middle terms as n = 9 is an odd number. The middle
terms will beth
2
19
and
th
2
39
terms.
t5 = 9C4(3x)517
43
x8
189
6
x
t6 = 9C5(3x)4 19
53
x1621
6x
.
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Illustration 8: Find the middle term in the expansion of
12a
bxx
.
Solution : 7th term is the middle term
T6+1 = 12C6 .6
x
a
. (b x)6
= 12C6 a6 b6
GREATEST BINOMIAL COEFFICIENTIn the binomial expansion of (1 + x)n , when n is even, the greatest binomial coefficient isgiven by nCn/2.
Similarly if n be odd, the greatest binomial coefficient will be
n 1 n 1
n n
2 2
C and C ,both being equal.+ -
NUMERICALY GREATEST TERMIf tr and tr + 1 be the r th and (r + 1)th term in the expansion of (1 + x)n, then
r
1r n
xC
xC
t
tr n
1r n
r r
n
r
1r
x.
Let numerically, tr + 1 be the greatest term in the above expansion. Then tr + 1 tr
or r
1r
t
t 1 r
1r n |x| 1
r |x|1
|x|1n
……(2)
Now shifting values of n and x in (2), we get r m + f or r mWhere m is a positive integer, f is a fraction such that 0 f < 1.
Now if f = 0 then tm + 1 and tm both the terms will be numerically equal and greatest while if f
0, then tm +1 is the greatest term of the binomial expansion.
i.e. to find the greatest term (numerically) in the expansion of (1 + x)n.
(i) Calculate m =|)x|1(
|x|)1n
.
(ii) If m is integer, then tm and tm + 1 are equal and are greatest term.
(iii) If m is not integer, then t[m] + 1 is the greatest term (where [.] denotes the greatest
integer function).
Illustration 9: Find the value of the greatest term in the expansion of 20
13 1
3
.
Solution: Since
r
20
r 20
r 1 r
r 1 20
r r 120
r 1
13 C
t C 1 21 r 13
t C r 3 313 C
3
r 1 r t t if only 21 – r r 3
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if only r 21 21( 3 1)
23 1
= 7.686
Hence t1 < t2 < t3 < t4 < t5 < t6 < t7 < t8 > t9 > t10
Hence t8 is the greatest term and its value is
7
20
7
13 C
3
=
20 20
7 76 3
1 1 258403 C C 2871.11
3 93
Illustration 10: Find numerically the greatest term in the expansion of ( ) 113 5x
when x =1
5.
Solution : Since11(3 5x) =
11
11 5x3 1
3
Now in the expansion of
115x
13
, we have
r 1
r
T (11 r 1) 5x
T r 3
=12 r 5 1
r 3 5
1x
5
=12 r 1
r 3
=12 r
3r
r 1
r
T 12 r 1 1
T 3r
4r 12
r 3 r 2,3
so, the greatest terms are 2 1T and 3 1T .
Greatest terms (where r = 2) = 11
2 13 | T |
=2
11 11
2
53 C x
3
=
2
11 11
2
5 13 C
3 5
1x
5
= 11 11.10 13
1.2 9 955 3
and greatest term (where r = 3) = 11
3 13 | T |
=
3
11 113 53 C x
3
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=
3
11 11
3
5 13 C
3 5
= 11 11.10.9 13
1.2.3 27
955 3
From above we say that the values of both greatest terms are equal.Alternative Method (Short Cut Method) :
Since
11 11
11 11 115x 1(3 5x) 3 1 3 1
3 3
1
x5
Now, calculate| x | (n 1)
m(| x | 1)
10
3
1(11 1)
3
1
13
= 3
The greatest terms in the expansion are 3T and 4
T
Greatest term (when r = 2) = 11
2 13 | T |
=
2
11 11
2
13 C
3
= 11 911.10 13 55 3
1.2 9
and greatest term (when r = 3) = 11
3 13 | T |
=
3
11 113
13 C3
= 11 11.10 9 13
1.2.3 27
955 3
From above we say that the values of both greatest terms are equal.
PROPERTIES OF BINOMIAL COEFFICIENT
For the sake of convenience the coefficientsn
C0 ,n
C1 , . . .,n
Cr , . . . ,n
Cn are usuallydenoted by C0, C1 , . . . , Cr , . . . ,Cn respectively
C0 + C1 + C2 +. . . . . + Cn = 2n
C0 - C1 + C2 -. . . . . + ( –1)n Cn = 0
C0 + C2 + C4 +. . . . . = C1 + C3 + C5 +. . . . . = 2n-1
21 r
nr
n CC r 1 = r 2 or r 1 + r 2 = n
nCr +nCr-1 = n+1Cr
r nCr =n n-1Cr-1
1n
C
1r
C 1r 1n
r n
.
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Illustration 11: Find the value of n
n
rr 0
r 2C
r 1
Solution: The given value is
n
0r
r n
n
0r
r n C
1r
11C
1r
2r
n
0r
1r 1n
n
0r
r n C
1n
1C )12(
1n
12 1nn
1n
1)3n(2n
Illustration 12: If (1 + x)n = C0 + C1x + C2x2 + . . . . . + Cnx
n,Show that (C0 + C1)(C1 + C2)(C2 + C3) . . . . . . (Cn-1 + Cn) =
( ).....
!
n
1 2 n
n 1C C C
n
Solution: As we know tr = Cr – 1 + Cr =n + 1Cr =
)!r 1n(!r
)!1n(
=)!1r n()!1r (
!n
r
)1n(
= 1r Cr
1n
Hence C0 + C1 = 0C1
1n
C1 + C2 = 1C2
1n
…… ……
Cn - 1 + Cn = 1nCn
1n
(C0 + C1)(C1 + C2) . . . . (Cn – 1 + Cn) = 1n10
n
C......CC!n
)1n(
PROBLEMS RELATED TO SERIES OF BINOMIAL COEFFICIENTS
Problems involving binomial coefficients with alternate sign:
Illustration 13: Evaluate C0 - C1 + C2 - C3 +...+ (-1)nCn.Solution: Here alternately +ve and - ve sign occur
This can be obtained by putting (-1) instead of 1 in place of x in(1 + x)n = C0 + C1x +...+ nCnx
n, we get C0 - C1 +...+ (-1)nCn = 0Now to obtain the sum C0 + C2 + C4 + ...we add (1 + 1)n and (1 - 1)n.Similarly, the cube roots of unity may be used to evaluateC0 + C3 + C6 + ... OR C1 + C4 +... OR C2 + C5 +...put x = 1, x = w, x = w2 in(1 + x)n = C0 + C1x +...+ Cnx
n and add to get C0 + C3 + C6 +...the other two may be obtained by suitably multiplying (1 + w)n and(1 + w2)n by w and w2 respectively.
Problems Related to series of Binomial coefficients in which
each term is a product of an integer and a binomial coefficient i.e.in the form k nCr
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Illustration 14: If (1+x)n =
n
0r
r
r xC then prove that C1 + 2C2 + 3C3+. . .+ nCn= n2n-1.
Solution: Method (i): By summation
r th term of the given series, tr = r nCr tr = n n-1Cr-1
Sum of the series =
n
1r
r t = 1r
n
1r
1n Cn
= 1n1n
11n
01n C.....CCn
= n 2n-1 .
Method (ii) By calculus
We have ( 1+ x )n = C0 + C1x + C2 x2 + . . . + Cnxn . . .(1)
Differentiating (1) with respect to xn(1 +x )n-1 = C1 +2C2x + 3C3 x2 + . . . + n Cnx
n-1 . . . (2)Putting x = 1 in (2), n 2n-1 = C1 + 2C2 + . . . + n nCn
Problems related to series of binomial coefficients in whicheach term is a binomial coefficient divided by an integer i.e. in
the form of n
r C
k.
Illustration 15: Prove that ....n
0 2 4C C C 2
1 3 5 n 1
Solution: Consider the expansion
(1 x)n C0 C1x C2x2 C3x
3 C4x4 … Cnx
n …(i)Integrating both sides of (i) within limits –1 to 1, we get
1 1n 2 3 4 n
0 1 2 3 4 n1 1(1 x) dx (C C x C x C x C X ...C x )dx
1 1
2 4 3
0 2 4 1 31 1(C C x C x ...)dx (C x C x ....)dx
1
2 4
0 2 41
2 (C C x C x ...)dx 0
(By Prop. Of definite integral)(since
second integral contains odd function)1
3 5n 11 2 4
1 0
0
C x C X(1 x)2 C x ...
n 1 3 5
n 12 4
0
C C22 C ...
n 1 3 5
Hencen
2 40
C C 2C ...
3 5 n 1
Alternative Method.
L.H.S. C0 2 4C C...
3 5
1 n(n 1) n(n 1)(n 2)(n 2)
....1.2.3 1.2.3.4.5
1 (n 1) (n 1)n(n 1) (n 1)n(n 1)(n 2)(n 3)
...(n 1) 1 1.2.3 1.2.3.4.5
1
(n 1)
{n1C1n1C3n1C5…}
1
(n 1){sum of even binomial coefficients of (1 x)n1}
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n 1 12
(n 1)
n2
n 1 R. H. S. coefficient divided by an integer i.e. in the form
of k
Cr n
.
Problem related to series of binomial coefficients in which eachterm is a product of two binomial coefficients.
Solution Process:(1) If difference of the lower suffixes of binomial coefficients in each term is same.
i.e. C1C3 C2C4 C3C5 …
Here 3 – 1 4 – 2 5 – 3 … 2Case I: If each term of series is positive then
(1 x)n C0 C1x C2x2 …. Cnx
n …(i) Interchanging 1 and x,
(x 1)n C0xn C1x
n –1 C2xn –2 … Cn …(ii)
Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on bothsides
Or
Replacing x by1
xin (i), then
n
1 2 n0 2 n
C C C11 C ...
x x x x
….(iii)
Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on bothsides.
Case II: If terms of the series alternately positive and negative then(1 –x)n C0 – C1 x C2x
2 - … ( –1)nCnxn …(i)
and (x 1)n C0xnC1x
n –1 C2xn –2 … Cn …(ii)
Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on bothsides.
Or
Replacing x by1
xin (i), then
n n
1 2 n0 2 n
C C ( 1) C11 C ...
x x x x
…(iii)
Then multiplying (i) and (iii) and equate the coefficient of suitable power of x on both
sides.
Illustration 16: If I is integral part of (2 + 3 )n and f is fraction part of (2 + 3 )n,
then prove that (I + f) (1 –f) = 1. Also prove that I is an odd Integer.
Solution: (2 + 3)n = I + f where I is an integer and 0 f < 1
Here note that (2 - 3)n (2 +3)n = (4 - 3)n = 1
Since (2 + 3)n (2 -3)n = 1 it is thus required to prove that
(2 - 3)n = 1 - f
but, (2 - 3)n + (2 +3)n = [2n - C1.2n - 1.3 + C22
n - 2..(3)2 - ...]
+ [2n + C1.2n - 1.3 + C22
n -2..(3)2 - ...]
= 2[2
n
+ C2.2
n - 2
.3+C42
n - 4
.3
2
+ ...] = even integer Now 0 < (2 - 3) < 1
0 < (2 - 3)n < 1
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if (2 - 3)n = f ', then I + f + f ' = Even
Now O f < 1 and 0 < f ' < 1 ……(1) Also I + f + f ' = Even integer f + f ' = integer ……(2) (1) and (2) imply that f + f ' = 1 ( since 0 < f + f ' < 2)
I is odd and f ' = 1 - f (I + f) (1 - f) = 1.
BINOMIAL THEOREM FOR ANY INDEX
(1+x)n = 1+ nx + 2x!2
)1n(n + . . . + r n(n 1) (n r 1)
x terms uptor!
Observations:
Expansion is valid only when –1 <x <1
General term of the series (1+x)-n = Tr+1 = (-1)r r!
1)r (n2)1)(nn(n
xr
General term of the series (1-x)-n = Tr+1 =r!
1)r (n2)1)(nn(n
xr
If first term is not 1, then make first term unity in the following way:
(a+ x)n = an(1+x/a)n if a
x< 1
IMPORTANT EXPANSIONS
(1+ x)-1 = 1- x +x2 –x3 + . . . + (-1)r xr +. . .
(1 - x)-1 = 1+ x +x2 +x3 + . . .+ xr + . . .
(1+ x)-2 = 1- 2x +3x2 –4x3+ . . .+ (-1)r (r+1)xr +. . .
(1 - x)-2 = 1+ 2x +3x2 +4x3 + . . .+ (r+1)xr +. . .
(1+x)-3 = 1- 3x +6x2 –10x3 +. . .+ (-1)r !r
)2r )(1r ( xr +. . .
(1-x)-3 = 1+ 3x +6x2 +10x3 + . . .+ !r
)2r )(1r ( xr +. . .
In general coefficient of xr in (1 – x) – n is n + r –1Cr .
(1 – x) –p/q = 1 +
2
q
x
!2
qpp
q
x
!1
p
+ …….
(1 + x) –p/q = 1 –
2
q
x
!2
qpp
q
x
!1
p
– …….
(1 + x)p/q = 1 +
2
q
x
!2
qpp
q
x
!1
p
+ …….
(1 – x)p/q = 1 –
2
q
x
!2
qpp
q
x
!1
p
– …….
Illustration 17: If –1 < x < 1, show that (1 –x)-2 = 1 + 2x + 3x2 + 4x3 + …..to .
Solution: We know that if n is a negative integer or fraction
(1+x)n= 1 +
to......x!4
3n2n1nnx
!3
2n1nnx
!2
1nnx
!1
n 432
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Provided –1 < x < 1Putting n = -2 and –x in place of x, we get
(1+x)2 = 1 +
to......x!3
22122x
!2
122x
!1
2 32
= 1 + 2x + 3x2 + 4x3 + … to .
Illustration 18: Find the square root of (99)1/2
correct to 4 places of decimal.
Solution: 2
1
99 = 2/1
2
1
100
111001100
=
2/1
100
11100
= 2
1
2
1
2
1
01.11001.1100
=
to........01.!2
12
1
2
1
01.!1
2
1
1102
= 10[1 –0.005 –0.0000125 + ……… to ] = 10 (.9949875) = 9.94987= 9.9499
MULTINOMIAL EXPANSION
In the expansion of (x1+x2 + . . . + xn)m where m, n N and x1, x2 , . . ., xn are independent
variables, we have
Total number of term in the expansion = m+n-1Cn-1
Coefficient of n321 r n
r 3
r 2
r 1 xxxx (where r 1 + r 2 +…+ r n = m, r i N {0} is
!r !r !r
!m
n21
.
Sum of all the coefficient is obtained by putting all the variables xi equal to 1 and it is
equal to nm.
Illustration 19: If x1 + x2 + x3 + x4 + x5 = 20 and x1 + x2 = 5 , (x1 ,x2 , x3 ,x4 , x5 0)then find the number of non negative integral solutions of aboveequation.
Solution: x1 + x2 + x3 + x4 + x5 = 20 , x1 + x2 = 5 … (1)
x3 + x4 + x5 = 15 … (2) Number of solutions
Coefficient of x5 in (1) coefficient of x15 in (2)
Coefficient of x5
26
x1
x1
Coefficient of x15
316
x1
x1
Coefficient of x5 in (1 – x)-2 Coefficient of x15 3)x1(
2 + 5 – 1C1 3+15-1C3-1 = 6C1 17C2 = 81617482
16176
3. OBJECTIVE ASSIGNMENTS
1: If roots of the equation m m m 20 1 mC C ....... C x - n n n
0 2 4C C C ....... x +
n n n1 3 5C C C ....... =0 are real , find minimum value of n – m.
(A) 1 (B) 2 (C) 3 (D) –1Solution: (C) Roots are real if
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02)2(42 1nm21n
022 1nm2n2
2n – 2 m+ n +1
n – m 3minimum value of n – m = 3
2: A number is said to be a nice number if it has exactly 4 factors.
(Including one and number itself). Let n = 23 32 53 7 112 , thennumber of factors, which are nice numbers is(A) 36 (B) 12 (C) 10 (D) 147
Solution: (B) Any number having exactly 4 factors is of the form m = p3 (p prime) or m = p.q (where p & q are distinct primes)So we have 5C2 + 2 = 12 such factors.
3: ( )nn
r 1 r
r 1
C1
r 1
is equal to
(A) –1
n 1(B) –
1
n
(C)1
n 1(D)
n
n 1
Solution: (D) Givennn
r 1 r
r 1
C( 1)
r 1
=n
r 1 n 1
r 1r 1
1( 1) C
n 1
=1
(0 1 (n 1))n 1
=n
n 1
4: ( )300
r 2 3 100
rr 0
a x 1 x x x
. If a =300
r
r 0
a
then300
rr 0
r a
is equal to
(A) 300 a (B) 100 a(C) 150 a (D) 75 a
Solution: (C) 300
r 2 3 100
r r 0
a x (1 x x x )
Clearly, ‘ar ’ is the coefficient of xr in the expansion of (1 + x + x2 + x3)100 .
Replacing x
1
x in the given equation, we getr 300
3 2 100
r 300r 0
1 1a (x x x 1)
x x
300
300 r 2 3 100
r r 0
a x (1 x x x )
Here ar represents of coefficient. of x300 –r in (1 + x + x2 + x3)100 Thus, ar = a300 –r
Let, I =300
r r 0
r a
=300
300 r r 0
(300 r)a
=300
r r 0
(300 r)a
=300 300
r r r 0 r 0
a r a
2I = 300 a
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I = 150 a
5: The number of terms in the expansion of
(1 + x)(1 + x3)( 1+ x6)( 1+x12) (1+x24) . . . . (1 +n23x ) is
(A) 2n+3 (B) 2n+4 (C) 2n+5 (D) none of these
Solution: (D) After expansion, no two terms will have the same powers of x or theterms are non over- lapping. Therefore, the total number of terms = 2 2 2
. . . (n +2) times = 2n+2 as a particular power of x can be chosen from eachbracket in 2 ways.
6: Number of terms in (1 +x)101 (1 + x2 –x)100 is(A) 302 (B) 301(C) 202 (D) 101
Solution: (C)(1 +x)101 (1 + x2 –x)100 = (1 +x) (1 + x3)100 = (1 + x) [C0 + C1x
3 + C2x6 + …..+ C100x
300 ]
= C0 + C0x + C1x3 + C1x4 + C2x6 + C2x7 + ……+ C100x300 + C100x301 Total number of terms = 101 + 101
= 202
7: If coefficient of x2 y3 z4 in (x + y +z)n is A, then coefficient of x4y4z is
(A) 2A (B)nA
2
(C)A
2(D) none of these
Solution: (C) Since x2y3z4 is occurring in the expansion of (x +y +z)n, so n should be 9 only.
Now A =!4!3!2
!9
= 1260
Coefficient of x4y4 z is!4!4
!9
= 630 = A/2.
8: Let n be an odd natural number and A =
n 1
2
nr 1 r
1
C
. Then value of
n
nr 1 r
r
Cis equal to
(A) n( A-1) (B) n( A+1)
(C)nA
2(D) nA
Solution: (B) Let n= 2m +1
A =1m1m2m2m21 C
1...
C
1
C
1
C
1...
C
1
C
1
2A +2 =
n
0r r C
1
Let S =
n
1r r C
r =
n
0r r C
r =
n
0r r nC
r n=
n
0r r C
r n.
2S = n
n
0r r C
1 S = n(A+1).
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9: Let rth term of a series be given by tr =2 4
r
1 3r r . Then
n
rn
r 1
lim t
is
(A) 3/2 (B) 1/2(C) -1/2 (D) –3/2
Solution: (C)
Tr can be written as
Tr = 222 r 1r
r
=
r 1r
1
r 1r
1
2
122
1r 22
1r
r r 1r
1
r 1r
1
2
1T
= 0........11551112
1
0
r 1r
1limas
2r
= – 2
1.
10: The coefficient of a4
b5
in the expansion of (a + b)9
is(A)
!5!4
!9(B)
!3!6
!9
(C)!9
!5!4(D) none of these
Solution: Coefficient of a4b5 will be!5!4
!9
Hence (A) is the correct answer.
11: The coefficient in the third term of the expansion of
n2
4
1
x
whenexpanded in decreasing powers of x is 31, then n is equal to(A) 16 (B) 20(C) 30 (D) 32
Solution: The third term will be nC2
2
4
1
= 31
162
1nn
= 31
n(n –1) = 32 . 31 n = 32.
Hence (D) is the correct answer.
12: The sum of coefficients in the expansion of (1 + x –3y2)2163 is
(A) 1 (B) –1(C) 22163 (D) none of these
Solution: For sum of coefficient put x = 1 and y = 1.
Hence (B) is the correct answer.
13: The sum of the rational terms in the expansion of 105/132 is
(A) 20 (B) 21(C) 40 (D) 41
Solution: There will be only two rational term the first term and the second term
2
5
+ 3
2
= 41.Hence (D) is the correct answer.
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14: If n is even then the coefficient of x in the expansion of (1 + x)nn
x
11
is
(A) nC2 (B) 2nCn (C) 0 (D) 1
Solution: (1 + x)n
n
x
1
1
= x –n(1 –x2)n
Since n is even only even power of x will occur in the expansion. Hence coefficient of
x is equal to zero.
Hence (C) is the correct answer.
15: The sum of 21C10 +21C9 + ……..+21C0 is equal to
(A) 220 (B) 221 (C) 219 (D) none of these
Solution: (1 + x)21 = 21C0 + 21C1x + 21C2x2 + …….+ 21C10 x10 + …….+ 21C21x
21
Put x = 1 (21C0 + 21C1 + 21C2 + …….+ 21C10) + (21C11 + …….+ 21C21) = 221
2(21C0 + 21C1 + ……..+ 21C10) = 221 21C0 + 21C1 + ……..+ 21C10 = 220.
Hence (A) is the correct answer.
16: In the expansion of 15
2
3
x
1x
, the constant term is
(A) 15C6 (B) –15C6 (C) 15C4 (D) –15C4
Solution: tr + 1 = ( –1)r 15Cr (x3)15 –r
r
2x
1
= ( –1)r 15Cr x
45 –3r –2r
For term independent of x 45 –5r = 0 r = 9 term independent of x will be = –15C9 = –15C6
Hence (B) is the correct answer.
17: 351 when divided by 8 leaves the remainder(A) 1 (B) 6(C) 5 (D) 3
Solution: 351 = 3.350 = 3(8 + 1)25
= 3(25C0825 + ……..+ 25C218) + 3
Hence (D) is the correct answer.
18: The greatest positive integer which divides n (n +1)(n + 2)(n + 3) for all n
N, is(A) 2 (B) 6(C) 24 (D) 120
Solution: Since product of r consecutive integer is divisible by r!
Hence (C) is the correct answer.
19: If 3
2
T
Tin the expansion of (a + b)n and
4
3
T
Tin the expansion of (a b)n +3 are
equal, then n is equal to(A) 3 (B) 4(C) 5 (D) 6
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Solution: 3n
33n
21n2
3n
22n2
n
1n1
n
baC
baC
baC
baC
6
)1n)(2n)(3n(2
)2n)(3n(
2
)1n(n
n
1n
3
1n
2
2n + 2 = 3n –3 n = 5
Hence (C) is the correct answer.20: The coefficient of xn in the expansion of (1 –x) –2 is
(A) ( –1)n(n + 1) (B) (n + 1)(C) ( –1)nn (D) none of these
Solution: Since (1 –x)2 = 1 + 2x + 3x2 + ………+ (n + 1)xn + ……
Hence (B) is the correct answer.
21: If n is a positive integer which of the following will always be integers?
I. ( 2 + 1)2n + ( 2 – 1)2n II. ( 2 + 1)2n –( 2 – 1)2n
III. ( 2 + 1)2n +1 + ( 2 – 1)2n +1 IV. ( 2 + 1)2n +1 –( 2 – 1)2n +1 (A) only I and III (B) only I and II(C) only I and IV (D) only II and III
Solution: In I and IV only even powers of 2 occurs whereas in II and III only odd powers of
2 occurs.
Hence (C) is the correct answer.
22: Coefficient of x5 in the expansion of (1 + x2)5(1 + x)4 is(A) 61 (B) 59(C) 0 (D) 60
Solution: (1 + x2)5(1 + x)4 = (1 + 5x2 + 10x4 + …..)(1 + x)4
Coefficient of x5 = 5 4C3 + 10 4C1 = 20 + 40 = 60.
Hence (D) is the correct answer.
23: The sixth term in the expansion of 8
102
3/8xlogx
x
1
is 5600 when x is
equal to(A) 10 (B) loge 10(C) 1 (D) none of these
Solution: T6 = 8C5 5
102
3
3/8 xlogxx
1 56x2(log10 x)5 = 5600
x2 (log10 x)5 = 100, obviously x = 10 satisfies the above equation.
Hence (A) is the correct answer.
24: The term independent of x in
10
2x2
3
3
x
is
(A) 1 (B) 5/12(C) 10C1 (D) None of these
Solution: General term in the expansion is
2
r 10
2
2
r
r
10
x2
3
3
x
C
=
2
r 10
r 5102
r 3
r
10
2
3
xC
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For constant term, 102
r 3
3
20r
which is not an integer. Therefore, there will be no constant term.Hence (D) is the correct answer.
25. If (1+ x + x2)n = a0 + a1x+ a2x2 + …+a2n x2n, then the value of a0 + a3 + a6 + . . . . is
(A) a1+ a4 +a7 + . . . (B) a1+ a2 +a3 + . . .(C) 2n +1 (D) none of these.
Solution: (1 + x + x2)n = a0 + a1 x + a2 x2 + a3 x3 + …put x = w, w2 we get0 = (a0 + a3 + a6 + …) + w (a1 + a4 + a7 + … ) + w2 (a2 + a5 + a8 + … ) … (1)0 = (a0 + a3 + a6 + …) + w2 (a1 + a4 + a7 + … ) + w (a2 + a5 + a8 + … ) … (2) from (1) and (2) we get,a0 + a3 + a6 … = a1 + a4 + a7 + …
26. The value of 2nCn - nC1.2n-2Cn +nC2 . 2n-4Cn – . . . . is equal to
(A) 3n (B) 4n
(C) 5n (D) none of these
Solution: 2nCn nC1
2n 2Cn + nC2 2n 4Cn …
= coefficient of xn in [n C0 (1 + x)2n n C1 (1 + x)2n2 + nC2 (1 + x)2n4 –….… ]
= coefficient of xn in [1 (1 + x)2]n = 2n
27. If | x | < 1, then the coefficient of xn in the expansion of (1 + x + x2 + x3 +…..)2 is
(A) n (B) n –1
(C) n + 2 (D) n + 1
Solution: (1 + x + x2 + x3 + … )2 =2
x1
1
= (1 x)2 = 1 + 2 x + 3 x2 + 4 x3 + …
coefficient of xn = (n + 1)
28. If (1+ax)n = 1+8x +24x2+……. then
(A) a= 3 (B) n= 5
(C) a= 2 (D) none of these
Solution: (1 + a x)n = 1 + n a x +2
)1n(n a2 x2 + … = 1 + 8 x + 24 x2 + … n a = 8
n (n 1) a2 = 48 n = 4, a = 2
29. The two successive terms in the expansion of (1+ x)24 whose coefficients are in the ratio 4 :1
are
(A) 3rd and 4th (B) 4th and 5th
(C) 5th and 6th (D) 6th and 7th
Solution: Let the coefficient of successive terms be 24Cr and 24Cr+1, then
1r 24
r 24
C
C
= 4 )r 24(
1r
= 4 r = 19
24C19,24C20 24C5,
24C4 6th and 5th terms
30. The co-efficient of xk (0 k n) in the expansion of E = 1+(1+x) +(1+x)2+ . . .(1+x)n is
(A) n+1Ck+1 (B) nCk
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(C) n+1Cn-k –1 (D) none of these
Solution: E =1)x1(
1)x1( 1n
=x
1...xCxCC 22
1n1
1n0
1n
= n+1C1 + n+1C2 x + n+1C3 x2 + …Coefficient of x4 = n+1Ck+1
31. The co-efficient of y in the expansion of (y2
+c/y)5
is(A) 10 c3 (B) 20 c2
(C) 10 c (D) 20 c
Solution: (r + 1)th terms = 5Cr y102r . Cr . yr
power of y = 1
10 3 r = 1 r = 3Required coefficient = 5C2 . x3 = 10 x3
32. If the coefficients of x2 and x3 in the expansion of (3 + kx)9 are equal, then thevalue of k is
(A)
9
7- (B)
9
7
(C)7
9(D) None of these.
Solution: r 1T
+in 9 9 9 r r
r (3 kx) C 3 (kx)-+ =
9 9 r r r
r C 3 k x-=
Coefficient of r 9 9 r r
r x C 3 k-= .
Now coefficient of x2 = coefficient of x3 9 9 2 2 9 9 3 3
2 3C 3 k C 3 k- -\ =
7 2 6 3
36 3 k 84 3 k
9
36 28k k7
.
Hence (B) is the correct answer.
33. The coefficient of xn in
2nn32
!n
x1....
!3
x
!2
xx1
is
(A)
!n
nn
(B)
!n
2n
(C) 2
!n1 (D) –
2!n1
Solution: Coefficient of xn in
2nn32
!n
x)1(...
!3
x
!2
xx1
Coefficient of xn in
232
...!3
x
!2
xx1
Coefficient of xn in (ex)2
Coefficient of xn in e2x =!n
)2( n
34. If C0, C1, C2, . . . . Cn are binomial coefficients then
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0
nn
2n
2
1nnn
C3
21......C
3
2C
3
2Clin is
(A) 0 (B) 1
(C) –1 (D) 2
Solution: Take x =
3
2
]xC)1(...xCxCC[lim n0
22n1nn
n
= ]xC)1(...xCxCC[lim nn
n2210
n
= n
n]x1[lim
=n
n 3
21lim
=
nn 3
1lim
= 0
35. Let n be an odd natural number and A =
2
1n
1r r n C
1. Then value of
n
1r r n C
r is equal to
(A) n(A – 1) (B) n( A + 1)
(C)2
nA(D) nA
Solution:
n
1r r n C
r =
1n
0r r n C
)r n(
=
1n
0r r n
1n
0r r n C
r
C
n
2
1n
0r r n
n
1r r n C
1n
C
r + n + n = n 2 A + 2 n
=
n
1r r n C
r = n (A + 1)
36. The sum of coefficients of even powers of x in the expansion of
11
x
1x
is
(A) 11 11C5 (B)2
11 11C6
(C) 11 611
511 CC (D) 0
Solution: (r + 1)th term = 11Cr (x)11r . x r
= 11Cr . x112r
Even power of x exists only if 11 2 r = an even number not possibleSum of coefficient = 0
37. The coefficient of xn in the expansion of x3x1
1
is
(A)1n
1n
3.2
13
(B)
1n
1n
3
13
(C)
1n
1n
3
13
2 (D) none of these.
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Solution:
x3
1
x1
1
2
1
)x3()x1(
1=
2
1[(1 x)1 (3 x)1]
=2
1
1
11
3
x13)x1(
=
...3
x
3
x
13
1
...xxx12
12
32
coefficient of xn =
n3
1
3
11
2
1=
1n
1n
32
13
38. If in the expansion of
x
x
4
12 , T3/T2 = 7 and the sum of the co-efficient of 2nd and 3rd
term is 36, then the value of x is
(A) –1/3 (B) –1/2
(C) 1/3 (D) ½
Solution: Given that nC1 + nC2 = 36
n = 8, n 9
also1
1nx1
n
2
2nx2
n
x4
1)2(C
x4
1)2(C
= 7
1
7x
2
6x
x41)2(8
x4
1)2(28
= 7 8x =
2
1 x =
3
1
39. The co-efficient of middle term in the expansion of (1+x)2n is
(A) 2 nCn (B) n2!n
)1n2(5.3.1
(C) 2. 6. ……(4n-2) (D) None of these
Solution: Coefficient of the middle term = 2nCn
=!n!n
)n2...4321(
=
!n
)1n2...531(2n
40. If 13Cr is denoted by Cr , then the value of C1+C5+C7+C9+C11 is equal to
(A) 212-287 (B) 212-165
(C) 212-C2-C13 (D) none of these
Solution: (1 + x)13 = C0 + C1 x + C2 x2 + … + C13 x13
(1 x)13 = C0 C1 x + C2 x2 … C13 x13 put x = 1213 = C0 + C1 + C2 … + C13
0 = (C0 + C2 + C4 + C6 + … ) (C1 + C3 + …)
2
13
= 2 (C0 + C2 + C4 + … C12 )212 = C0 + C2 + C4 + … C12 L.H.S. = C1 + C5 + C7 +C9 + C11 = C1 + C2 + C4 +C6 + C8
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= 212 1 13C10
= 212 287
41. If P(n) is a statement such that truth of P(n) the truth of P(n + 1) for n N,then P(n) is true(A) for all n
(B) for all n > 1(C) for all n > m, m is some fixed positive integer (D) nothing can be said.
Solution: Nothing can be said about the truth of P(n), for all n N because truth of P(1)is not given.Hence (D) is the correct answer.
42. If x > –1, then the statement P(n) : (1 + x)n > 1 + nx is true for
(A) all n N (B) all n > 1
(C) all n > a and x 0 (D) None of these.Solution: P(1) is not true.
For n = 2, P(2) : (1 + x)2 > 1 + 2x is true if x 0Let P(k) : (1 + x)k > 1 + kx be true
(1 + x)k+1 = (1 + x) (1 + x)k > (1 + x) (1 + kx)= 1 + (k + 1)x + kx2 > 1 + (k + 1)x ( kx2 > 0)
Hence (C) is the correct answer.
43. The greatest positive integer, which divides (n + 16) (n + 17) (n + 18) (n + 19),
for all n N, is
(A) 2 (B) 4(C) 24 (D) 120Solution: Since product of any r consecutive integers is divisible by r ! and not by (r+1)!
The given product is divisible by 4 ! = 24.Hence (C) is the correct answer.
44. A student was asked to prove a statement by induction. He proved (i) P(5) is
true and (ii) truth of P(n) truth of P(n + 1), n N. On the basis of this, hecould conclude that P(n) is true
(A) for no n (B) for all n 5(C) for all n (D) None of these.
Solution: Obviously (B) is the answer.
45. The inequality n ! > 2n –1 is true(A) for all n > 1 (B) for all n > 2
(C) for all n N (D) for no n NSolution: It is not true for n = 1, 2
For n > 2, n ! > 1 . 2 . 3 ......... (n – 1)n
> 2n –1 ( 2 2, 3 > 2, 4 > 2, ......., n > 2)Hence (B) is the correct answer.
46. The smallest positive integer for which the statement 3n+1 < 4n holds is(A) 1 (B) 2(C) 3 (D) 4
Solution: The given statement is true for n 4.
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Hence (D) is the correct answer.
47. 23n – 7n – 1 is divisible by(A) 64 (B) 36(C) 49 (D) 25
Solution: For n = 1, 23n – 7n – 1 has value 23 – 7 – 1 = 0
For n = 2, 2
3n
– 7n – 1 has value 26 – 14 – 1 = 49.which is divisible by 49 and not by 36 or 64.Hence (C) is the correct answer.
48. For each n N, 23n – 1 is divisible by(A) 8 (B) 16(C) 32 (D) None of these.
Solution: For n 1, 23n – 1 = (23)n – 1 = 8n – 1= (8 – 1) [8n –1 + 8n –2 + ....... + 1]
= 7 positive integer Hence (D) is the correct answer.
49. If the ratio of the 7th term from the beginning to the 7th term from the end in
the expansion of
x
3
3
12
3is
1
6, then x is
(A) 9 (B) 6(C) 12 (D) None of these.
Solution: 7T in
x 6
x 1/ 3 x 636 1/ 33
1 12 C (2 )
33
7th term from the end in
x3
3
12
3
= T7 in
x x 6
x 1/ 3 636 1/ 33
1 12 C (2 )
33
6
x 1/ 2 x 6
6 1/ 3
x 6
x 1/ 3 6
6 1/ 3
1C (2 )
13
61C (2 )
3
x 12 x 121/3 x 123 3
x 12
1/ 3
(2 ) 1 12 3
6 61
3
x 12
13x 12
6 6 1 x 93
.
Hence (A) is the correct answer.
50. If n 2 n
0 1 2 n(1 x) C C x C x ...... C x+ = + + + + , then
1 2 n
0
C C C
C ......2 3 n 1+ + + +
+ is equal to
(A) n 12 + (B)n 12 1
n 1
+
-
+
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(C)n 12
n 1
+
+(D) n 12 1+
-
Solution: 1 2 n0
C C CC ......
2 3 n 1+ + + +
+
n n(n 1)1
1 1.21 .....2 3 n 1
-
= + + + + +
n n(n 1) 11 ......
2! 3! n 1
-= + + + +
+
1 (n 1)n (n 1) n(n 1)(n 1) ... 1
n 1 2! 3!
n 1 n 1 n 1 n 1
1 2 3 n 1
1C C C ... C
n 1
n 1n 1 n 1
0
1 2 12 C
n 1 n 1.
Hence (D) is the correct answer.
51. The term independent of x in the expansion of
61
2x3x
is
(A)160
9(B)
80
9
(C)160
27(D)
80
3
Solution:
r 6 r 6 6 r 6 6 2r
r 1 r r r
1 2
T C (2x) C x3x 3
Let r 1T
+be independent of x.
6 – 2r = 0 or r = 36 3
6 6 2(3)
r 1 3 1 3 3
2T T C x
3
-
-
+ +\ = =
20 8 160
27 27
´= =
Hence (C) is the correct answer.
52. The middle term in the expansion of 2n(1 x)+ is
(A) 2n
nC (B) 2n n 1
n 1C x +
+
(C) 2n n 1
n 1C x -
-(D) n n1.3.5... (2n 1)
2 xn!
-×
Solution: 2n is even.
Middle term2n 2n n n 2n n
2n 2 n 1 n n
2
T T C 1 x C x-
+ += = = =
n n2n! 1.2.3.4.5.6......2nx x
n! n! n! n!
= =
nn1.3.5.......(2n 1) 2 .n!
xn!n!
-=
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Hence (D) is the correct answer.
53. If the binomial expansion of 2(a bx)-+ is
13x ......
4- + , where a > 0, then (a, b)
is(A) (2, 12) (B) (2, 8)
(C) ( –2, 12) (D) None of these.
Solution:
2
2 2 b(a bx) a 1 x
a
2 2 3
1 b 1 2b1 ( 2) x ...... x .....
a a a a
Also, 2 1(a bx) 3x .....
4
-+ = - +
2
1 1
a 4\ = . . . (1) and
3
2b3
a- = - . . . (2)
(1) 2a 4 a 2 and from (2) b = 12
Hence (A) is the correct answer.
54. 2 1/ 2(4 5x )-- can be expanded as a power series of x if
(A) | x | 5 / 2< (B) | x | 2 / 5<
(C) – 1 < x < 1 (D) None of these.
Solution:
1/ 2
2 1/ 2 1/ 2 25(4 5x ) 4 1 x
4
1/ 221 5
1 x2 4
.
25x 1
4\ - < or
2 25x | x | 1
4- < or 25
x 14
<
or 2 4x
5< or
2| x |
5< .
Hence (B) is the correct answer.
55. In the binomial expansion of (a – b)n, n 5, the sum of the 5th and 6th terms
is zero. The a/b is equal to
(A)n 5
6
-(B)
n 4
5
-
(C)5
n 4-
(D)6
n 5-
Solution: n n 4 4 n n 5 5
5 6 4 5T T 0 C a ( b) C a ( b) 0 n n 4 4 n n 5 5
4 5C a b C a b
n
5
n
4
Ca n! 4!(n 4)! n 4
b C 5!(n 5)! n! 5.
Hence (B) is the correct answer.56. If the coefficient of mth, (m + 1)th and (m + 2)th terms in the expansion
n(1 x)+ are in A.P., then
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(A) 2 2n 4(4m 1) 4m 2 0+ + + - =
(B) 2 2n n(4m 1) 4m 2 0+ + + + =
(C) 2(n 2m) n 2- = +
(D) 2(n 2m) n 2+ = +
Solution: We have n n n
m 1 m m 1C , C , C- +
in A.P.
n n nm m 1 m 12 C C C
2(n!) n ! n!
m!(n m)! (m 1)! (n m 1)! (m 1)! (n m 1)!
2 1 1
m (n m) (n m 1) (n m) m(m 1)
2(m 1) (n m 1) m(m 1) (n m 1)(n m)
On simplification, we get
2 2 2n 4mn 4m n 2 0 (n 2m) n 2 .
Hence (C) is the correct answer.
57. If 2 n 2 2n
0 1 2 2n(1 x x ) a a x a x ...... a x- + = + + + + , then
0 1 2 2na a a ...... a+ + + + is equal to
(A)n3 1
2
+(B)
n3 1
2
-
(C) n 13
2- (D) n 1
32
+
Solution: We have2 n 2 2n
0 1 2 2n(1 x x ) a a x a x ....... a x- + = + + + +
Putting x = 1 and – 1, we get
0 1 2 3 2n1 a a a a ...... a= + + + + +
and n
0 1 2 3 2n3 a a a a ...... a= - + - + +
Adding, we getn
0 2 4 2n1 3 2(a a a ...... a )+ = + + + + n
0 2 4 2n
3 1a a a ...... a
2
+\ + + + + = .
Hence (A) is the correct answer.
58. The positive value of a so that the coefficients of x5
and x15
are equal in the
expansion of
10
2
3
ax
x
(A)1
2 3(B)
1
3
(C) 1 (D) 2 3
Solution:
r
10 2 10 r 10 r 20 5r
r 1 r r 3
aT C (x ) C a x
x
20 5r 5 r 3 10 3 20 5(3) 3 5
r 1 3 1 3T T C a x 120a x-
+ +\ = = =
Coefficient of x5 = 120 a3
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Also, 20 5r 15 r 1 10 1 20 5(1) 15
r 1 1 1 1T T C a x 10ax-
+ +\ = = =
Coefficient of x15 = 10a
3120a 10a\ = or 1
a2 3
= .
Hence (A) is the correct answer.
59. The term independent of x in the expansion of n n(1 x) (1 1/ x)+ + is
(A) 2 2 2 2
0 1 2 nC 2C 3C ...... (n 1)C+ + + + +
(B) 2
0 1 n(C C ...... C )+ + +
(C) 2 2 2
0 1 nC C ...... C+ + +
(D) None of these.Solution: We have
n
n 2 n 1 2 n0 1 2 n 0 2 nC C C1(1 x) 1 (C C x C x ...... C x ) C ......
x x x x
Term independent of x on the R.H.S.2 2 2 2
0 1 2 nC C C ...... C= + + + + .
Hence (C) is the correct answer.
60. The coefficient of x3 in
6
5
3
3x
xis
(A) 0 (B) 120
(C) 420 (D) 540
Solution:
r
6 5 / 2 6 r
r 1 r 3 / 2
3T C (x )
x
5r 3 r 15
6 r 6 r 15 4r 2 2r r C 3 x C 3 x
- --
= =
Let r 1T+
contains x3.
15 4r 3\ - = or r = 36 3 15 4(3)
r 1 3 1 3T T C (3) x -
+ +\ = =
3 320 27 x 540x= ´ ´ =
Coefficient of x
3
= 540.Hence (D) is the correct answer.
61. In the expansion of 50(1 x)+ , let S be the sum of coefficients of odd power of
x, then S is(A) 0 (B) 249 (C) 250 (D) 251
Solution:50 50 50 2 50 3 50 49 50 50
1 2 3 49 50(1 x) 1 C x C x C x ...... C x C x+ = + + + + + +
Sum of coefficients of odd powers of x50 50 50 50 1 49
1 3 49C C ...... C 2 2-+ + + = = .
Hence (B) is the correct answer.
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62. The coefficient of x53 in100
100 100 r r
r r
C (x 3) 2--å is
(A) 100
51C (B) 100
52C
(C) 100
53C- (D) 100
54C
Solution:100
100 100 r r
r r 0
C (x 3) 2-
=
-
å
100 100 100((x 3) 2) (x 1) (1 x)= - + = - = - 100 100
100 r r r 100 r
r r r 0 r 0
C ( x) ( 1) ( 1) C x= =
- = - -å å
Coefficient of 53 53 100 100
53 53x ( 1) C C= - = - .
Hence (C) is the correct answer.