BINOMIAL THEOREM-JEE(MAIN)

RESONANCE JEE(MAIN) BINOMIAL THEOREM - 1

Binomial expression :

Any algebraic expression which contains two dissimilar terms is called binomial expression.

For example : x + y, x2y + 2xy

1, 3 � x, 1x2 + 3/13 )1x(

1

etc.

Terminology used in binomial theorem :

Factorial notation : or n! is pronounced as factorial n and is defined as

n! =

0nif;1

Nnif;1.2.3)........2n)(1n(n

Note : n! = n . (n � 1)! ; n N

Mathematical meaning of nCr : The term nC

r denotes number of combinations of r things choosen

from n distinct things mathematically, nCr =

!r)!rn(!n

, n N, r W, 0 r n

Note : Other symbols of of nCr are

r

n and C(n, r).

Properties related to nCr :

(i) nCr = nC

n � r

Note : If nCx = nC

y Either x = y or x + y = n

(ii) nCr + nC

r � 1 = n + 1C

r

(iii)1r

nr

n

C

C

= r1rn

(iv) nCr =

rn

n�1Cr�1

= )1r(r)1n(n

n�2C

r�2 = ............. = 1.2).......2r)(1r(r

))1r(n().........2n)(1n(n

(v) If n and r are relatively prime, then nCr is divisible by n. But converse is not necessarily true.

Statement of binomial theorem :

(a + b)n = nC0 anb0 + nC

1 an�1 b1 + nC

2 an�2 b2 +...+ nC

r an�r br +...... + nC

n a0 bn

where n N

or (a + b)n =

n

0r

rrnr

n baC

Note : If we put a = 1 and b = x in the above binomial expansion, thenor (1 + x)n = nC

0 + nC

1 x + nC

2 x2 +... + nC

r xr +...+ nC

n xn

or (1 + x)n =

n

0r

rr

n xC

Binomial Theorem

�Obvious� is the most dangerous word in mathematics......... Bell, Eric Temple


Regarding Pascal�s Triangle, we note the following :(a) Each row of the triangle begins with 1 and ends with 1.(b) Any entry in a row is the sum of two entries in the preceding row, one on the immediate left and

the other on the immediate right.

Example # 3 : The number of dissimilar terms in the expansion of (1 � 3x + 3x2 � x3)20 is(A) 21 (B) 31 (C) 41 (D) 61

Solution : (1 � 3x + 3x2 � x3)20 = [(1 � x)3]20 = (1 � x)60

Therefore number of dissimilar terms in the expansion of (1 � 3x + 3x2 � x3)20 is 61.

General term :

(x + y)n = nC0 xn y0 + nC

1 xn�1 y1 + ...........+ nC

r xn�r yr + ..........+ nC

n x0 yn

(r + 1)th term is called general term and denoted by Tr+1

.T

r+1 = nC

r xn�r yr

Note : The rth term from the end is equal to the (n � r + 2)th term from the begining, i.e. nCn � r + 1

xr � 1 yn � r + 1

Example # 4 : Find (i) 28th term of (5x + 8y)30 (ii) 7th term of 9

2x5

54x

Solution : (i) T27 + 1

= 30C27

(5x)30� 27 (8y)27 = !27!3

!30 (5x)3 . (8y)27

(ii) 7th term of

9

x25

5x4

T6 + 1

= 9C6

69

5x4

6

x25

= !6!3

!9

3

5x4

6

x25

= 3x

10500

Example # 5 : Find the number of rational terms in the expansion of (91/4 + 81/6)1000.

Solution : The general term in the expansion of 10006/14/1 89 is

Tr+1

= 1000Cr

r1000

41

9

r

61

8

= 1000Cr 2

r1000

3

2r

2

The above term will be rational if exponent of 3 and 2 are integers

It means 2

r1000 and

2r

must be integers

The possible set of values of r is {0, 2, 4, ............, 1000}Hence, number of rational terms is 501

Middle term(s) :

(a) If n is even, there is only one middle term, which is

th

22n

term.

(b) If n is odd, there are two middle terms, which are th

21n

and

th

12

1n

terms.

Example # 6 : Find the middle term(s) in the expansion of

(i)

142

2x

1

(ii)

93

6a

a3


Case - When

ba

1

1n

is an integer (say m), then

(i) Tr+1

> Tr

when r < m (r = 1, 2, 3 ...., m � 1)i.e. T

2 > T

1, T

3 > T

2, ......., T

m > T

m�1

(ii) Tr+1

= Tr

when r = mi.e. T

m+1 = T

m

(iii) Tr+1

< Tr

when r > m (r = m + 1, m + 2, ..........n )i.e. T

m+2 < T

m+1 , T

m+3 < T

m+2 , ..........T

n+1 < T

n

Conclusion :

When ba

1

1n

is an integer, say m, then TTm and T

m+1 will be numerically greatest terms (both terms are

equal in magnitude)Case -

When

ba

1

1n

is not an integer (Let its integral part be m), then

(i) Tr+1

> Tr

when r <

ba

1

1n

(r = 1, 2, 3,........, m�1, m)

i.e. T2 > T

1 , T

3 > T

2, .............., T

m+1 > T

m

(ii) Tr+1

< Tr

when r >

ba

1

1n

(r = m + 1, m + 2, ..............n)

i.e. Tm+2

< Tm+1

, Tm+3

< Tm+2

, .............., Tn +1

< Tn

Conclusion :

When

ba

1

1n

is not an integer and its integral part is m, then T

m+1 will be the numerically greatest

term.

Note : (i) In any binomial expansion, the middle term(s) has greatest binomial coefficient.In the expansion of (a + b)n

If n No. of greatest binomial coefficient Greatest binomial coefficientEven 1 nC

n/2

Odd 2 nC(n � 1)/2

and nC(n + 1)/2

(Values of both these coefficients are equal )(ii) In order to obtain the term having numerically greatest coefficient, put a = b = 1, and proceed

as discussed above.

Example # 8 : Find the numerically greatest term in the expansion of (3 � 5x)15 when x = 51

.

Solution : Let rth and (r + 1)th be two consecutive terms in the expansion of (3 � 5x)15

Tr + 1

Tr

15Cr 315 � r (| � 5x|)r 15C

r � 1 315 � (r � 1) (|� 5x|)r � 1

!r!)r15()!15

|� 5x | !)1r(!)r16(

)!15.3

5 . 51

(16 � r) 3r

16 � r 3r4r 16 r 4


Self practice problems :

(8) If n is a positive integer, then show that 32n + 1 + 2n + 2 is divisible by 7.

(9) What is the remainder when 7103 is divided by 25 .

(10) Find the last digit, last two digits and last three digits of the number (81)25.

(11) Which number is larger (1.2)4000 or 800

Answers : (9) 18 (10) 1, 01, 001 (11) (1.2)4000.

Some standard expansions :(i) Consider the expansion

(x + y)n =

n

0rr

nC xn�r yr = nC0 xn y0 + nC

1 xn�1 y1 + ...........+ nC

r xn�r yr + ..........+ nC

n x0 yn ....(i)

(ii) Now replace y � y we get

(x � y)n =

n

0rr

nC (� 1) r xn�r yr = nC0 xn y0 � nC

1 xn�1 y1 + ...+ nC

r (�1)r xn�r yr + ...+ nC

n (� 1)n x0 yn ....(ii)

(iii) Adding (i) & (ii), we get(x + y)n + (x � y)n = 2[nC

0 xn y0 + nC

2 xn � 2 y2 +.........]

(iv) Subtracting (ii) from (i), we get(x + y)n � (x � y)n = 2[nC

1 xn � 1 y1 + nC

3 xn � 3 y3 +.........]

Properties of binomial coefficients :

(1 + x)n = C0 + C

1x + C

2x2 + ......... + C

r xr + .......... + C

nxn ......(1)

where Cr denotes nC

r

(1) The sum of the binomial coefficients in the expansion of (1 + x)n is 2n

Putting x = 1 in (1)

nC0 + nC

1 + nC

2 + ........+ nC

n = 2n ......(2)

or

n

0r

nr

n 2C

(2) Again putting x = �1 in (1), we get

nC0 � nC

1 + nC

2 � nC

3 + ............. + (�1)n nC

n = 0 ......(3)

or

n

0rr

nr 0C)1(

(3) The sum of the binomial coefficients at odd position is equal to the sum of the binomial coefficientsat even position and each is equal to 2n�1.from (2) and (3)

nC0 + nC

2 + nC

4 + ................ = nC

1 + nC

3 + nC

5 + ................ = 2n�1

(4) Sum of two consecutive binomial coefficients

nCr + nC

r�1 = n+1C

r L.H.S. = nC

r + nC

r�1=

!r)!rn(!n

+

)!1r()!1rn(!n

= )!1r()!rn(!n

1rn

1

r

1 = )!1r()!rn(

!n

)1rn(r)1n(

= !r)!1rn(

)!1n(

= n+1C

r = R.H.S.


Method : By Integration

(1 + x)n = C0 + C

1x + C

2x2 + ...... + C

n xn.

Integrating both sides, within the limits � 1 to 0.

0

1

1n

1n)x1(

=

0

1

1n

n

3

2

2

10 1nx

C.....3x

C2x

CxC

1n1

� 0 = 0 �

1n

C)1(.....

3

C

2

CC n1n21

0

C0 �

2C1 +

3C2

� .......... + (� 1)n 1n

Cn

=

1n1

Proved

Example # 14 : If (1 + x)n = C0 + C

1x + C

2x2 + ........+ C

nxn, then prove that

(i) C0

2 + C1

2 + C2

2 + ...... + Cn

2 = 2nCn

(ii) C0C

2 + C

1C

3 + C

2C

4 + .......... + C

n � 2 C

n = 2nC

n � 2 or 2nC

n + 2

(iii) 1. C0

2 + 3 . C1

2 + 5. C2

2 + ......... + (2n + 1) . Cn

2 . = 2n. 2n � 1Cn + 2nC

n.

Solution : (i) (1 + x)n = C0 + C

1x + C

2x2 + ......... + C

n xn. ........(i)

(x + 1)n = C0xn + C

1xn � 1+ C

2xn � 2 + ....... + C

n x0 ........(ii)

Multiplying (i) and (ii)(C

0 + C

1x + C

2x2 + ......... + C

nxn) (C

0xn + C

1xn � 1 + ......... + C

nx0) = (1 + x)2n

Comparing coefficient of xn,C

02 + C

12 + C

22 + ........ + C

n2 = 2nC

n

(ii) From the product of (i) and (ii) comparing coefficients of xn � 2 or xn + 2 both sides,C

0C

2 + C

1C

3 + C

2C

4 + ........ + C

n � 2 C

n = 2nC

n � 2 or 2nC

n + 2.

(iii) Method : By Summation

L.H.S. = 1. C0

2 + 3. C1

2 + 5. C2

2 + .......... + (2n + 1) Cn

2.

=

n

0r

)1r2( nCr2 =

n

0r

r.2 . (nCr)2 +

n

0r

2r

n )C(

= 2

n

1r

n. . n � 1Cr � 1

nCr + 2nC

n

(1 + x)n = nC0 + nC

1 x + nC

2 x2 + .............nC

n xn ..........(i)

(x + 1)n � 1 = n � 1C0 xn � 1 + n � 1C

1 xn � 2 + .........+n � 1C

n � 1x0 .........(ii)

Multiplying (i) and (ii) and comparing coeffcients of xn.n � 1C

0 . nC

1 + n � 1C

1 . nC

2 + ........... + n � 1C

n � 1 . nC

n = 2n � 1C

n

n

0r1r

1n C . nCr = 2n � 1C

n

Hence, required summation is 2n. 2n � 1Cn + 2nC

n = R.H.S.

Method : By Differentiation

(1 + x2)n = C0 + C

1x2 + C

2x4 + C

3x6 + ..............+ C

n x2n

Multiplying both sides by xx(1 + x2)n = C

0x + C

1x3 + C

2x5 + ............. + C

nx2n + 1.

Differentiating both sidesx . n (1 + x2)n � 1 . 2x + (1 + x2)n = C

0 + 3. C

1x2 + 5. C

2 x4 + .....+ (2n + 1) C

n x2n......(i)

(x2 + 1)n = C0 x2n + C

1 x2n � 2 + C

2 x2n � 4 + ......... + C

n........(ii)

Multiplying (i) & (ii)(C

0 + 3C

1x2 + 5C

2x4 + ......... + (2n + 1) C

n x2n) (C

0 x2n + C

1x2n � 2 + ........... + C

n)

= 2n x2 (1 + x2)2n � 1 + (1 + x2)2n

comparing coefficient of x2n,C

02 + 3C

12 + 5C

22 + .........+ (2n + 1) C

n2= 2n . 2n � 1C

n � 1 + 2nC

n.

C0

2 + 3C1

2 + 5C2

2 + .........+ (2n + 1) Cn

2= 2n . 2n�1Cn + 2nC

n. Proved


Multinomial theorem :

As we know the Binomial Theorem �

(x + y)n =

n

0rr

nC xn�r yr =

n

0r!r)!rn(

!n xn�r yr

putting n � r = r1 , r = r

2

therefore, (x + y)n = nrr 21

21!r!r

!n 21 rr y.x

Total number of terms in the expansion of (x + y)n is equal to number of non-negative integral solutionof r

1 + r

2 = n i.e. n+2�1C

2�1 = n+1C

1 = n + 1

In the same fashion we can write the multinomial theorem

(x1 + x

2 + x

3 + ........... x

k)n =

nr...rr k21k21

!r!...r!r!n

k21 r

kr2

r1 x...x.x

Here total number of terms in the expansion of (x1 + x

2 + .......... + x

k)n is equal to number of non-

negative integral solution of r1 + r

2 + ........ + r

k = n i.e. n+k�1C

k�1

Example # 17 : Find the coefficient of a2 b3 c4 d in the expansion of (a � b � c + d)10

Solution : (a � b � c + d)10 = 10rrrr 4321

4321!r!r!r!r

)!10( 4321 rrrr )d()c()b()a(

we want to get a2 b3 c4 d this implies that r1 = 2, r

2 = 3, r

3 = 4, r

4 = 1

coeff. of a2 b3 c4 d is !1!4!3!2)!10(

(�1)3 (�1)4 = � 12600

Example # 18 : In the expansion of 11

x7

x1

, find the term independent of x.

Solution :11

x7

x1

=

11rrr 321321

!r!r!r)!11(

3

21

rrr

x7

)x()1(

The exponent 11 is to be divided among the base variables 1, x and x7

in such a way so that we

get x0.Therefore, possible set of values of (r

1, r

2, r

3) are (11, 0, 0), (9, 1, 1), (7, 2, 2), (5, 3, 3), (3, 4, 4),

(1, 5, 5)Hence the required term is

)!11()!11(

(70) + !1!1!9)!11(

71 + !2!2!7)!11(

72 + !3!3!5)!11(

73 + !4!4!3)!11(

74 + !5!5!1)!11(

75

= 1 + !2!9)!11(

. !1!1!2

71 + !4!7)!11(

. !2!2!4

72 + !6!5!)11(

. !3!3!6

73

+ !8!3!)11(

. !4!4!8

74 + !10!1!)11(

. !5!5!)10(

75

= 1 + 11C2 . 2C

1 . 71 + 11C

4 . 4C

2 . 72 + 11C

6 . 6C

3 . 73 + 11C

8 . 8C

4 . 74 + 11C

10 . 10C

5 . 75

= 1 +

5

1rr2

11C . 2rCr . 7r


Example-20 : If x is so small such that its square and higher powers may be neglected, then find the value of

2/1

3/52/1

)x4(

)x1()x31(

Solution : 2/1

3/52/1

)x4(

)x1()x31(

= 2/1

4x

12

3x5

1x23

1

= 21

x

619

2

2/1

4x

1

= 21

x

619

2

8x

1 = 21

x

619

4x

2 = 1 � 8x

� 1219

x = 1 � 24

41 x

Self practice problems :(16) Find the possible set of values of x for which expansion of (3 � 2x)1/2 is valid in ascending

powers of x.

(17) If y = 52

+ !23.1

2

52

+

!35.3.1

3

52

+ ............., then find the value of y2 + 2y

(18) The coefficient of x100 in 2)x1(

x53

is

(A) 100 (B) �57 (C) �197 (D) 53

Answers : (16) x

23

,23

(17) 4 (18) C


20. Find the coefficient of the term independent of x in the expansion of 10

2/13/13/2 xx

1x

1xx

1x

21. If in the expansion of (1 + x)m (1 � x)n, the coefficients of x and x2 are 3 and �6 respectively. Then find the value

of m.

22. Find the number of terms in the expansion of (1 + 5 2 x)9 + (1 � 5 2 x)9.

23. If the coefficients of second, third and fourth terms in the expansion of (1 + x)n are in A.P., then find the valueof n.

24. If in the expansion of (1 � x)2n�1 the coefficient of xr is denoted by ar, then prove that a

r�1 + a

2n�r = 0

25. Using binomial theorem, prove that 23n � 7n � 1 is divisible by 49 where n N.

26. Using binomial theorem, prove that 32n+2 � 8n � 9 is divisible by 64, n N.

27. Prove that 5.4

1�

4.31

3.21

�2.1

1 + ..... = loge

e4

.

28. Find the sum of the infinite series 1 + !61

!41

!21

+ ........

29. Prove that (x2 � y2) + !21

(x4 � y4) + !31

(x6 � y6) + ...... to = 22 yx e�e

Type (IV) : Very Long Answer Type Questions: [06 Mark Each]

30. Find the value of

n

0rrn

e

er

nr

)10log1(

10logr1C)1( .

31. If the coefficient of rth, (r + 1)thand (r + 2)th terms in the expansion of (1 + x)14 are in A.P, then findthe value of r.

32. If the coefficients of three cosecutive terms in the expansion of (1 + x)n are in the ratio 1 : 7 : 42. Find n

33. If 3rd, 4th, 5th and 6th terms in the expansion of (x + )n be respectively a, b, c and d then prove that

bdc

acb2

2

=

c3a5

34. If coefficients of three consecutive terms in the expansion of (1 + x)n be 76,95 and 76. Then find n.

35. If the 2nd, 3rd and 4th terms in the expansion of (x + a)n are 240, 720 and 1080 respectively, find x, a and n.

36. Sum the series from n = 1 to n = , whose nth term is

(i) !)1n(1

(ii) !)2n(1

(iii) !)1�n2(1

37. Prove that

loge

nm

= 2

....

nmn�m

51

nmn�m

31

nmn�m

53

38. Prove that

loge

x1x

=

....

)1x2(5

1

)1x2(3

1)1x2(

12

53


A-12. The co-efficient of x in the expansion of (1 2 x3 + 3 x5) 1

18

x is :

(1) 56 (2) 65 (3) 154 (4) 62

A-13. The term containing x in the expansion of

52

x1

x

is -

(1) 2nd (2) 3rd (3) 4th (4) 5th

A-14. Given that the term of the expansion (x1/3 x1/2)15 which does not contain x is 5 m, where m N,then m=(1) 1100 (2) 1010 (3) 1001 (4) none

A-15. The term independent of x in the expansion of 34

x1

xx1

x

is:

(1) 3 (2) 0 (3) 1 (4) 3

A-16. The term independent of x in the expansion of

10

2x2

33x

is-

(1) 3/2 (2) 5/4 (3) 5/2 (4) None of these

A-17. Let the co-efficients of xn in (1 + x)2n & (1 + x)2n 1 be P & Q respectively, then5

QQP

=

(1) 9 (2) 27 (3) 81 (4) none of these

A-18. If (1 + by)n = (1+ 8y + 24 y2 +....) where nN then the value of b and n are respectively-(1) 4, 2 (2) 2, � 4 (3) 2, 4 (4) � 2, 4

A-19. The coefficient of x52 in the expansion

100

0mm

100C (x � 3)100�m. 2m is :

(1) 100C47

(2) 100C48

(3) �100C52

(4) �100C100

A-20. The co-efficient of x5 in the expansion of (1 + x)21 + (1 + x)22 +....... + (1 + x)30 is :(1) 51C

5(2) 9C

5(3) 31C

6 21C

6(4) 30C

5 + 20C

5

A-21. The term independent of x in (1 + x)m

n

x1

1

is

(1) m � nCn

(2) m + nCn

(3) m + 1Cn

(4) m + nCn+1

A-22. (1 + x) (1 + x + x2) (1 + x + x2 + x3)...... (1 + x + x2 +...... + x100) when written in the ascending powerof x then the highest exponent of x is(1) 5000 (2) 5030 (3) 5050 (4) 5040

Section (B) : Numerically greatest term, Remainder and Divisibility problems

B-1. The numerically greatest term in the expansion of (2 + 3 x)9, when x = 3/2 is(1) 9C6. 29. (3/2)12 (2) 9C3. 29. (3/2)6 (3) 9C5. 29. (3/2)10 (4) 9C4. 29. (3/2)8

B-2. The numerically greatest term in the expansion of (2x + 5y)34, when x = 3 & y = 2 is :(1) T21 (2) T22 (3) T23 (4) T24

B-3. The remainder when 22003 is divided by 17 is :(1) 1 (2) 2 (3) 8 (4) none of these


C-9. The value of

0

50

1

50 +

1

50

2

50 +...........+

49

50

50

50 is, where nC

r =

r

n

(1)

50

100(2)

51

100(3)

25

50(4)

2

25

50

C-10. The value of

10

1r 1rn

rn

C

C.r is equal to

(1) 5 (2n � 9) (2) 10 n (3) 9 (n � 4) (4) none of these

C-11. The value of the expression

10

0rr

10 C

10

0KK

K10

K

2

C)1( is :

(1) 210 (2) 220 (3) 1 (4) 25

C-12. In the expansion of (1 + x)n

n

x1

1

, the term independent of x is-

(1) 20C + 2 2

1C +.....+ (n + 1) 2nC (2) (C0 + C1 +....+ Cn)2

(3) 20C + 2

1C +.......+ 2nC (4) None of these

C-13. If (1 + x)n = C0 + C1x + C2x2 +...+Cn.xn then for n odd, C12 + C3

2 + C52 +.....+ Cn

2 is equal to

(1) 22n � 2 (2) 2n (3) 2)!n(2

)!n2((4) 2)!n(

)!n2(

C-14. If an =

n

0r rnC

1, the value of

n

0r rnC

r2n is :

(1) 2n

an

(2)41

an

(3) nan

(4) 0

Section (D) : Multinomial Theorem, Binomial Theorem for negative and fractional index

D-1. The coefficient of a5 b4 c7 in the expansion of (bc + ca ab)8 is(1) 280 (2) 240 (3) 180 (4) 32

D-2. If x < 1, then the co-efficient of xn in the expansion of (1 + x + x2 + x3 +.......)2 is(1) n (2) n 1 (3) n + 2 (4) n + 1

D-3 The coefficient of x4 in the expression (1 + 2x + 3x2 + 4x3 + ......up to )1/2 (where | x | < 1) is(1) 1 (2) 3 (3) 2 (4) 5

Section (E) : Exponential and Logarithmic series

E-1_. Sum of the infinite series

!4321

!321

!21

+ ..... to

(1) 3e

(2) e (3) 2e

(4) none of these

E-2_. The coefficient of x6 in series e2x is

(1) 454

(2) 453

(3) 452

(4) none of these


4. In the expansion of

20

43

6

14

(1) the number of irrational terms is 19 (2) middle term is irrational(3) the number of rational terms is 2 (4) All of these

5. If (1 + 2x + 3x2)10 = a0 + a1x + a2x2 +.... + a20x20, then :(1) a1 = 20 (2) a2 = 210 (3) a4 = 8085 (4) All of these

6. (1 + x + x2 + x3)5 = a0 + a

1x + a

2x2 +....................... + a

15x15, then a

10 equals to :

(1) 99 (2) 101 (3) 100 (4) 110

7. In the expansion of

n

23

x

1x

, n N, if the sum of the coefficients of x5 and x10 is 0, then n is :

(1) 25 (2) 20 (3) 15 (4) None of these

8. The coefficient of the term independent of x in the expansion of

10

21

31

32

xx

1x

1xx

1x

is :

(1) 70 (2) 112 (3) 105 (4) 210

9. The term in the expansion of (2x � 5)6 which has greatest binomial coefficient is(1) T3 (2) T4 (3) T5 (4) T6

10. The remainder when 798 is divided by 5 is(1) 4 (2) 0 (3) 2 (4) 3

11. The last three digits of the number (27)27 is(1) 805 (2) 301 (3) 503 (4) 803

12. 79 + 97 is divisible by :(1) 7 (2) 24 (3) 64 (4) 72

13. Let f(n) = 10n + 3.4n +2 + 5, n N. The greatest value of the integer which divides f(n) for all n is :(1) 27 (2) 9 (3) 3 (4) None of these

14. Coefficient of xn 1 in the expansion of, (x + 3)n + (x + 3)n 1 (x + 2) + (x + 3)n 2 (x + 2)2 +..... + (x + 2)n

is :(1) n+1C2(3) (2) n1C2(5) (3) n+1C2(5) (4) nC2(5)

15. The term in the expansion of (2x � 5)6 which has greatest numerical coefficient is(1) T3 ,T4 (2) T4 (3) T5 , T6 (4) T6 , T7

16. Number of elements in set of value of r for which, 18Cr 2 + 2. 18Cr 1 + 18Cr 20C13 is satisfied :(1) 4 elements (2) 5 elements (3) 7 elements (4) 10 elements

17. The number of values of ' r

' satisfying the equation, 2r

391r3

39 CC = r3

391r

39 CC 2

is :

(1) 1 (2) 2 (3) 3 (4) 4

18. The sum 1

1 1

1

2 2

1

1 1! ( ) ! ! ( ) !......

! ( ) !n n n

is equal to :

(1)1

n ! (2n 1 1) (2)

2

n ! (2n 1) (3)

2

n ! (2n1 1) (4) none


PART - II : COMPREHENSION

Comprehension # 1

Let P be a product given by P = (x + a1) (x + a

2) ......... (x + a

n)

and Let S1 = a

1 + a

2 + ....... + a

n =

n

1iia , S

2 =

jiji ,a.a S

3 =

kjikji a.a.a and so on,

then it can be shown thatP = xn + S

1 xn � 1 + S

2 xn � 2 + ......... + S

n.

1. The coefficient of x8 in the expression (2 + x)2 (3 + x)3 (4 + x)4 must be(1) 26 (2) 27 (3) 28 (4) 29

2. The coefficient of x19 in the expression (x � 1) (x � 22) (x � 32) .......... (x � 202) must be(1) 2870 (2) 2800 (3) �2870 (4) � 4100

3. The coefficient of x98 in the expression of (x � 1) (x � 2) ......... (x � 100) must be

(1) 12 + 22 + 32 + ....... + 1002

(2) (1 + 2 + 3 + ....... + 100)2 � (12 + 22 + 32 + ....... + 1002)

(3) 21

[(1 + 2 + 3 + ....... + 100)2 � (12 + 22 + 32 + ....... + 1002)]

(4) None of these

Comprehension # 2

We know that if nC0, nC

1, nC

2, ........., nC

n be binomial coefficients, then (1 + x)n = C

0 + C

1 x + C

2 x2 + C

3x3

+ ......+ Cn xn. Various relations among binomial coefficients can be derived by putting

x = 1, � 1, i,

23i

21

,1iwhere .

4. The value of nC0 � nC

2 + nC

4 � nC

6 + ....... must be

(1) 2i (2) (1 � i)n � (1 + i)n

(3) 21

[(1 � i)n + (1 + i)n] (4) 21

[(2 � i)n + (1 � i)n]

5. The value of expression (nC0 � nC

2 + nC

4 � nC

6 + .......)2 + (nC

1 � nC

3 + nC

5 .........)2 must be

(1) 22n (2) 2n (3) 2n2 (4) None of these

PART - I : AIEEE PROBLEMS (LAST 10 YEARS)

1. If rnC denotes the number of combinations of n things taken r at a time, then the expression

rn

1rn

1rn C2CC

equals [AIEEE 2003]

(1) r2n C (2) 1r

2n C (3) r

1n C (4) 1r1n C

2. The number of integral terms in the expansion of 2568 53 is : [AIEEE 2003]

(1) 32 (2) 33 (3) 34 (4) 35.


13. The sum of the series 20C0 � 20C

1 + 20C

2 � 20C

3 + ..... + 20C

10 is [AIEEE 2007 (3, �1), 120]

(1) �20C10

(2) 21

20C10

(3) 0 (4) 20C10

14. In the binomial expansion of (a � b)n , n 5, the sum of 5th and 6th term is zero, then ba

equals

[AIEEE 2008 (3, �1), 105]

(1) 5

4n (2)

4n5

(3) 5n

6

(4) 6

5n

15. Statement-1 :

n

0r

)1r( nCr = (n + 2) 2n�1 [AIEEE 2008 (3, �1), 105]

Statement-2 :

n

0r

(r + 1) nCr xr = (1 + x)n + nx (1 + x)n � 1

(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(3) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is True

16. Let S1 =

10

1j

)1�j(j 10Cj , S

2 =

10

1j

j 10Cj and S

3 =

10

1j

2j 10Cj. [AIEEE 2009 (4, �1), 144]

Statement -1 : S3 = 55 × 29 .Statement -2 : S1 = 90 × 28 and S2 = 10 × 28.

(1) Statement -1 is true, Statement-2 is true ; Statement -2 is not a correct explanation for Statement -1.

(2) Statement-1 is true, Statement-2 is false.

(3) Statement -1 is false, Statement -2 is true.

(4) Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for Statement-1.

17. The coefficient of x7 in the expansion of (1 � x � x2 + x3)6 is : [AIEEE 2011 (4, �1), 120]

(1) 144 (2) � 132 (3) � 144 (4) 132

18. If n is a positive integer, then n213 � n2

13 is : [AIEEE-2012, (4, �1)/120]

(1) an irrational number (2) an odd positive integer(3) an even positive integer (4) a rational number other than positive integers

19. The term independent of x in expansion of

10

2/13/13/2 xx

1x

1xx

1x

is : [AIEEE - 2013, (4, � ¼) 120 ]

(1) 4 (2) 120 (3) 210 (4) 310


BOARD LEVEL SOLUTIONS

Type (I)

1. º

430

4

x4

xC

+

133

14

x4

)x(C

+

223

24

x4

)x(C

+ 3

133

4

x4

)x(C

+

403

44

x4

)x(C

x12 + 16 x8 + 96 x4 + 256 + 4x

256 Ans.

2. 06 C (ax)6

º

xb

+ 6C

1 (ax)5

1

xb

�

+ 6C2 (ax)4

2

xb

�

+ 6C

3 (ax)3

3

xb

�

+ 6C4 (ax)2

4

xb

�

+ 6C

5 ax

5

xb

�

+ 6C6 (ax)6

6

xb

�

= a6x6 � 6a5bx4 + 15a4b2x2 � 20a3b3

+ 2

42

x

ba15 � 4

5

x

ab6 + 6

6

x

b

3.

04

04

xa

ax

C

+

13

14

xa

ax

C

+

22

24

xa

ax

C

+

31

34

xa

ax

C

+

40

44

xa

ax

C

= 2

2

a

x �

ax4

+ 6 � 4xa

+ 2

2

x

a Ans.

4. e2x+3

= e2x.e3 = e3

...

!3)x2(

!2)x2(

!1)x2(

132

Thus the coefficient of x2 in the expansion of

e2x+3 is e3

!222

= 2e3

5. We have, ex = 1 + !4

x!3

x!2

x!1

x 432

+ .... to

Putting x = 2, we get

e2 = 1 +!7

2!6

2!5

2!4

2!3

2!2

2!1

2 765432

+...

e2 = 1 + 2 + 2 + 1·333 + 0.666 + 0.266

+ 0.088 + 0.025 e2 = 7·378

e2 = 7·4 (correct to one decimal place)

6. loge(1 + 3x + 2x2) = loge[(1 + 2x) (1 + x)]= loge(1 + 2x) + loge(1 + x)

=

...)x2(

41

�)x2(31

2)x2(

�x2 432

+

...)x(

41

�)x(31

)x(21

�x 432

= 3x �25

x2 + 3x3 �4

17x4 + ...

Type (II)

7. 05 C (1 + x)5 (�x2)0 + 1

5C (1 + x)4 (�x2)1

+ 25 C (1 + x)3 (�x2)2

+ 35 C (1 + x)2 (�x2)3 + 4

5C (1 + x)1 (�x2)4

+ 55 C (1 + x)0 (�x2)5

(1 + 5x + 10x2 + 10x3 + 5x4 + x5)+ 5[1 + 4x + 6x2 + 4x3 + x4](�x2)+ 10[1 + 3x + 3x2 + x3] (x4)+ 10[1 + x2 + 2x] (�x6) + 5(1 + x) (x8) + (�x10)

(1 + 5x + 10x2 + 10x3 + 5x4 + x5)+ [�5x2 � 20x3 � 30 x4 � 20x5 � 5x6]+ (10x4 + 30x5 + 30x6 + 10x7)+ [�10 x6 � 10x8 � 20x7] + 5x8 + 5x9 � x10

�x10 + 5x9 � 5x8 � 10x7 + 15x6 + 11x5

� 15x4 � 10x3 + 5x2 + 5x + 1 Ans.

8. 03 C (x + 2)3

º

x1

+ 1

3C (x + 2)2

1

x1

+ 23 C (x + 2)1

2

x1

+ 3

3 C (x + 2)0

3

x1

[x3 + 8 + 12x + 6x2] + 3.[x2 + 4x + 4].

x1

+ 3(x + 2).

2x

1 � 3x

1


16. As Tr+1

= rrnr

n yxC in (x + y)n

Now consider

17

x7

x

[on comparing n = 17, r = 10, x = x , y = x7

]

T11

= 1017C (x)17�10

10

x7

= 10

107

1017

x

)7(.xC

T11

= 710 310

17 xC Ans.

Type (III)

17. (101)100 = (1 + 100)100

[(1 + x)n = 0nC + 1

nC x + 22

n xC +.......+ nn

n xC ]

Using Binomial theorem

(1 + 100)100 = 0100C + 1

100C (100)1

+ 2100C (100)2 +....... + 100

100C (100)100

Now (1+ 100)100 � 1 = 1 + 104 + 2100C 104 +...10200 � 1

= 104 [1 + 2100C +....... 10196]

1 + 2100C +..... + 10196 is a natural number by the

virtue of its being the binomial coefficients.= 104 N

(101)100 � 1 is divisible by 10,000.

18. Consider 171995 + 111995 � 71995

(7 + 10)1995 + (1 + 10)1995 � 71995

01995C (7)1995 + 1994

11995 )7(C (10)1

+ .... 19951995C (10)1995 + 0

1995C + 11995C (10)1

+..... 19951995C (10)1995 � 71995

Now 01995C + 10

[ 19941

1995 )7(C +........ 19941995

1995 )10(C

+ 11995C +......... 1995

1995C (10)1994]

1 + 10N [ 11995C (7)1994 +.......

19951995C (10)1994 + 1

1995C +........ 19951995C (10)1994]

= N(natural number as it is the sum of binomialcoefficients) Units place is 1 Ans.

19. Consider

7

21x4

21

[ (x + y)n = 0

nC xn +

1nC xn�1 y1 + 2

nC xn�2 y2 +....... + nnC yn]

= 7

07

21

C

+

6

17

21

C

21x4

+

25

27

21x4

21

C

+.....+

7

77

21x4

C

...(i)

Now

7

21x4

21

= 7

07

21

C

+

6

17

21

C

21x4

+

25

27

21x4

21

C

+...+

7

77

21x4

C

...(ii)

(i) � (ii)

= 2[

6

17

21

C

21x4

+

34

37

21x4

21

C

+

52

57

21x4

21

C

+

7º

77

21x4

21

C

]

= 2 1x4 [ 717

2

1.C + 73

7

2

1.C . (4x + 1)

+ 757

2

1.C (4x + 1)2 +

27)1x4( 3

]

= 62

11x4 [ 1

7C + 37 C (4x + 1)

+ 57 C (4x + 1)2 + (4x + 1)3]

1x4

1

77

21x41

21x41

= 62

1[ 1

7C + 37 C (4x + 1) + 5

7 C (4x + 1)2 + (4x + 1)3]

It is a polynomial of degree 3. Ans.

20. Let = x = t6

10

36

6

24

6

tt

1t

1tt

1t

10

33

33

24

242

)1t(t

)1t()1t(

1tt

)1tt()1t(

10

3

335

t

1ttt


= !)1n(!)1n2(!)1n2(

(�1)r �1 + !)rn2(!)1r(

!)1n2(

(�1)2n�r

= !)1r(!)rn2(!)1n2(

[(�1)r�1 + (�1)2n�r]

= !)1r(!)rn2(!)1n2(

r

r

)1(

111

= !)1r(!)rn2(!)1n2(

[0] = 0 proved.

25. Consider 23n � 7n � 1 = (8)n � 7n � 1

= (1 + 7)n � 7n � 1

[ (1 + x)n = 0nC + xC1

n + ....... + nn

n xC ]

= 1n7)7(C......)7(C)7(CC nn

n22

n11

n0

n

= 17n(7)C...(7)C(7)C7n1 nn

n33

n22

n

= nn

n33

n22

n )7(C......)7(C)7(C

= 2nn

n3

n2

n2 7C......7CC[7 ]

= 49[ 2nn

n3

n2

n 7C......7CC ]

2nn

n3

n2

n 7C......7.CC = N

It is a natural number by the virtue of being a sum ofbinomial coefficients.23n � 7n � 1 = 49 N

23n � 7n � 1 is divisible by 49. Proved.

26. Consider32n+2 � 8n � 9 = (32)n+1 � 8n � 9 = (9)n+1 � 8n � 9

= (1 + 8)n+1 � 8n � 9 ... same

= 9n8)8(C.......)8(C)8(CC 1n1n

1n22

1n11

1n0

1n

= 1 + (n + 1) 8 + 9n88.......)8(C 1n22

1n

= 1 + 8n + 8 + n+1C2 (8)2 + .......+ 8n+1 � 8n � 9

= n+1C2 (8)2 + n+1C

3 (8)3 + ....... + 8n+1

= (8)2 [n+1C2 + n+1C

3 (8) + ........ + 8n�1 ]

n+1C2 + n+1C

3 (8) + ...... + 8n�1 = N

It is a natural number by the virtue of being a sum ofbinomial coefficients. 32n+2 � 8n � 9 = 64N

32n+2 � 8n � 9 is divisible by 64

27. L.H.S. = 5.4

1�

4.31

3.21

�2.1

1 + ..... to

=

51

�41

�41

�31

31

�21

�21

�1 + ......

= 1 � 51

51

41

�41

�31

31

21

�21

..........

= 1 � 2

to.....

51

�41

31

�21

= 2

to......

51

41

�31

21

�1 � 1

= 2 loge 2 � 1 = loge4 � logee = loge

e4

= R.H.S.

28. We have ex = 1 + !3

x!2

x!1

x 32

+..... to

Put x = 1, we get e = 1 + !31

!21

!11

+ ....

Put x = � 1, we get e�1 = 1 � !31

�!2

1!1

1 +.......

add both equation, we get

e + e�1 =

.....

!61

!41

!21

12

Hence !61

!41

!21

1 + .... = 21

(e + e�1)

29. L.H.S. =

......

!4x

!3x

!2x

x864

2

�

......

!3y

!2y

y64

2

=

...!3)x(

!2)x(

x1322

2

�

...

!3)y(

!2)y(

y13222

2

= 22 yx e�e

Type (IV)30. Let log

e 10 = x

Now

n

0rrr

nr

)xn1(

xr1C)1( [ log am = m log a]

n

0rrr

nr

)xn1(

1C)1( +

n

0rrr

nr

)nx1(

rxC)1(

n

0rrr

nr

)nx1(

rC)1(

+ nx1

x

n

1r

r

rn

)1( 1r1n C

1r)nx1(

r

[using rnC =

rn

1r1n C ]


Now viv

= bdc

acb2

2

= 88n2

66n2

x

x

]C.C)C[(

]C.C)C[(

5n

3n2

4n

4n

2n2

3n

!5!)5n(!n

.!3)!3n(

!n!4!4!)4n(!)4n(

!n!n

!4!)4n(!n

!2!)2n(!n

!3!3!)3n(!)3n(!n!n

x2

2

!4!31

)3n(51

4)4n(1

!)5n(!)4n(!n!n

)2n(41

3)3n(1

!)4n(!)3n(!n!n

!3!21

x2

2

16n415n5)4n()3n(5

)3n).(2n(4.39n38n4

!)3n(!)5n(

!2!4

.x

2

2

)1n()4n(5.4

)2n.(12)1n(

!)3n(!)5n(

!2!4

.x

2

2

= 2

2x

20 !)2n(

!)4n(

34. Let Tr , T

r+1 and T

r+2 be the three consecutive terms in

the expansion of (1 + x)n

As Tr+1

= rr

n xC in [1 + x]n

Tr = 1r

1rn xC

T

r+1 = r

rn xC

Tr+2

= 1r1r

n xC

Now it is given that coefficients ofT

r , T

r+1 and T

r+2 are 76, 95, 76 repsectively.

1rnC

= 76 ...(i)

rnC = 95 ...(ii)

1rnC

= 76 ...(iii)

Now iii

= 1r

nr

n

C

C

= r

1rn =

7695

76n � 76r + 76 = 95 r

76(n + 1) = 101 r ...(iv)

iiiii

= r

n1r

n

C

C =

1rrn

=

9576

95n � 95r = 76r + 76

95n � 76 = 101r ...(v)From (iv) 95n � 76 = 76n + 76

19n = 152n = 8 Ans.

35. As Tr+1

= rnC xn�r yr in (x + y)n & consider (x + a)n

T2 = 1

nC xn�1 a1 = 240 (given) ...(i)

T3 = 2

nC xn�2 a2 = 720 ...(ii)

T4 = 3

nC xn�3 a3 = 1080 ...(iii)

iii

1

n2

n

C

C. 1n

2n

x

x

1

2

a

a = 3

= n2

)1n(n .

xa

= 3 (n � 1).xa

= 6 ...(iv)

iiiii

2

n3

n

C

C. 2n

3n

x

x

. 2

3

a

a =

7201080

= 23

)1n(n62).2n()1n(n

xa

= 23

(n � 2) xa

= 29

...(v)

viv

2n1n

=

96

.2 = 34

3n � 3 = 4n � 8

n = 5From (iv)

4.xa

= 6 xa

= 23

a = 2x3

Put a = 2x3

in (i) 1nC xn�1 a1 = 240

5.x4.2x3

= 240

x5 = 32 x = 2

Now a = 2x3

= 3

n = 5; a = 3; x = 2 Ans.

36.

(i) Sum =

1nnt = t1 + t2 + t3 + ..... to

=

1n!)1n(

1 = !4

1!3

1!2

1 + ..... to

=

2�.....

!31

!21

!11

1 = e � 2

(ii) We have, tn = !)2n(1

Sum =

1n!)2n(

1


OBJECTIVE QUESTIONS

* Marked Questions may have more than one correct option.

1. If the sum of the co-efficients in the expansion of (1 + 2x)n is 6561, then the greatest term in the expansionfor x = 1/2 is :(1) 4th (2) 5th (3) 6th (4) none of these

2. The expression,

6

22

622

1x21x2

21x21x2

is a polynomial of degree

(1) 5 (2) 6 (3) 7 (4) 8

3. Co-efficient of x5 in the expansion of (1 + x2)5 (1 + x)4 is :(1) 40 (2) 50 (3) 30 (4) 60

4. Co-efficient of x15 in (1 + x +x3 + x4)n is :

(1)

5

0rr

nr315

n CC (2)

5

0rr5

nC (3)

5

0rr3

nC (4)

3

0rr5

nr3

n CC

5. If n is even natural and coefficient of xr in the expansion of

x1x1 n

is 2n, (|x| < 1), then �

(1) 2/nr (2) 2/)2n(r (3) 2/)2n(r (4) nr

6. The coefficient of xn in polynomial (x + 2n+1C0) (x + 2n+1C

1)........(x + 2n+1C

n) is -

(1) 2n + 1 (2) 22n+1 � 1 (3) 22n (4) none of these

7.

n

1r

1r

0p

pp

rr

n 2CC is equal to -

(1) 4n � 3n + 1 (2) 4n � 3n � 1 (3) 4n � 3n + 2 (4) 4n � 3n

8. nC0 � 2.3 nC

1 + 3.32 nC

2 � 4.33 nC

3 +..........+ (�1)n (n +1) nC

n 3n is equal to

(1)

1

2n3

21 nn(2)

23

n2n(3) 2n + 5n 2n (4) (�2)n.

9. If the sum of the coefficients in the expansion of (2 + 3cx + c2x2)12 vanishes, then c equals to(1) �1, 2 (2) 1, 2 (3) 1, �2 (4) �1, �2

10. The term independent of x in the expansion of ( 1 + x + 2x2)

4

22

x3

1x3

is

(1) 10 (2) 2 (3) 0 (4) 6

11*. Let !n

1000a

n

n for n N, then an is greatest, when

(1) n = 997 (2) n = 998 (3) n = 999 (4) n = 1000

12. 2k

0

n

k

n � 1k2

1

n

1k

1n + 2k2

2

n

2k

2n �...... + (� 1)k

k

n

0

kn =

(1) nCk (2) n+1Ck (3) n�1Ck (4) n+2Ck

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