Unit 6 Binomial Theorem

28
 SYLLABUS Mathematical induction and its applications, Binomial theorem for a positive integral index, properties of binomial coefficients, 1. MA THEMA TICAL INDUCTION AND ITS APPLICA TIONS 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 proceed as 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. Illus tra tion 1: P rove tha t if sin   0, then n 1 n n 1 sin2 cos cos2 cos4 ...... cos2 2 sin + + a a × a × a a = a , holds for each n n. Solution: If P(n) denotes the given statement, then for n = 1, P(1 ): sin4 cos cos2 4sin a a × a = a , which is true because sin4 2sin2 cos2 4sin 4 sin 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 sin2 cos cos2 ...... cos2 2 sin + + a a × 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 sin2 cos cos2 ...... cos2 cos2 2 sin + + + a a × a a × a = a  L.H.S. m 1 m 1 m 1 sin2 sin cos2 2 sin + + + a = × a a  m 1 m 1 m 1 2sin2 cos 2 2 2 sin + + + a = × a  m 1 m 2 sin( 2 2 ) 2 sin + + × a = × a R.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

Transcript of Unit 6 Binomial Theorem

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 1/28

 

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=

= 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 =

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 =

L.H.S.m 1

m 1

m 1

sin2 sincos2

2 sin

+

+

+

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

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 2/28

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

i.e. (a – x)n =

n

0r

rrnr

nrxaC1  

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 3/28

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

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

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 4/28

are equal.

Solution : The general term in

11 r 

2 11 2 11 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

 

  

  .

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 5/28

Illustration 8: Find the middle term in the expansion of 

12a

bxx

.

Solution : 7th term is the middle term

T6+1 = 12C6 .6

x

  

 . (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

1r n

xC

xC

t

tr n

1r n

r r 

n

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

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  

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 6/28

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

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

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

 

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 7/28

=

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

.

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 8/28

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

)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  

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 9/28

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}

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 10/28

   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

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 11/28

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

  

 

 

  

  + …….

(1 + x) –p/q = 1 – 

2

q

x

!2

qpp

q

x

!1

  

 

 

  

   – ……. 

(1 + x)p/q = 1 +

2

q

x

!2

qpp

q

x

!1

  

 

 

  

  + ……. 

(1 – x)p/q = 1 – 

2

q

x

!2

qpp

q

x

!1

  

 

 

  

   – ……. 

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  

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 12/28

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 

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 13/28

  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

(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

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 14/28

  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

(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

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).

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 15/28

 

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 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

  

 = 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.

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 16/28

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  

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

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 17/28

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

10

x2

3

3

x

C

 

 

 

 

 

 

 

 =

2

r 10

r 5102

r 3

10

2

3

xC

 

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 18/28

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 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 

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 19/28

(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

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 20/28

 

 

  

 

 

  

 

 

  

 

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

]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

C

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.

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 21/28

Solution: 

x3

1

x1

1

2

1

)x3()x1(

1=

2

1[(1 x)1  (3 x)1]

=2

 

  

 

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

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 

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 22/28

= 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.

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 23/28

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

= 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

+

-

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 24/28

(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

(C)160

27(D)

80

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!

-=  

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 25/28

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

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

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

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 26/28

(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

(A)1

2 3(B)

1

(C) 1 (D) 2 3  

Solution:

10 2 10 r 10 r 20 5r  

r 1 r r  3

aT C (x ) C a 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 

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 27/28

 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:

6 5 / 2 6 r  

r 1 r  3 / 2

3T C (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.

7/29/2019 Unit 6 Binomial Theorem

http://slidepdf.com/reader/full/unit-6-binomial-theorem 28/28

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.