# binomial-jeemain

```JEEMAIN.GURU
JEE-Mathematics
BINOMIAL THEOREM
1.
BINOMIAL EXPRESSION :
Any algebraic expression which contains two dissimilar terms is called binomial expression.
For example : x – y, xy +
2.
1 1
1
,  1,
 3 etc.
x z
(x  y )1 / 3
BINOMIAL THEOREM :
The formula by which any positive integral power of a binomial expression can be expanded in the form of a
series is known as BINOMIAL THEOREM.
n
If x, y  R and n  N, then : ( x + y)n = nC0xn + nC1xn-1 y + nC2xn-2 y2 + ..... + nCrxn-r yr + ..... + nCnyn =

n
C r x n r y r
r 0
This theorem can be proved by induction.
Observations :
(a)
The number of terms in the expansion is ( n+1) i.e. one more than the index.
(b)
The sum of the indices of x &amp; y in each term is n.
(c)
The binomial coefficients of the terms (nC0, nC1.....) equidistant from the beginning and the end are equal.
i.e. n C r = n C r –1
(d)
n 
Symbol nCr can also be denoted by   , C(n, r) or A nr .
r 
Some important expansions :
(i)
(1 + x) n = n C 0 + n C 1 x + n C 2 x 2 + ........ + nC nx n .
(ii)
(1 – x) n = n C 0 – nC 1x + n C 2 x 2 + ........ + (–1) n . n C n x n .
Note :
The coefficient of x r in (1+x) n = nC r &amp; that in (1–x) n = (–1) r . nC r
Illustration 1 :
Expand : (y + 2)6.
Solution :
6C
0y
6
+ 6C1y5.2 + 6C2y4.22 + 6C3y3.23 + 6C4y2. 24 + 6C5y1 . 25 +
6C
6
. 26.
= y6 + 12y5 + 60y4 + 160y3 + 240y2 + 192y + 64.
E
Illustration 2 :

2y 2
Write first 4 terms of  1 
5

Solution :
 2y 2
7C , 7C  
0
1
 5
Illustration 3 :
The value of
 7
 2y 2
 , C 2  
 5


2
 7  2y 2
 , C 3  

 5
18
3



3
 7 3  3.18.7.25

is -
6
3  6.243.2  15.81.4  20.27.8  15.9.16  6.3.32  64
(A) 1
Solution :
7



(B) 2
(C) 3
The numerator is of the form a 3  b 3  3ab  a  b    a  b 
Where, a = 18 and b = 7
Denominator can be written as

(D) 4
3
Nr = (18 + 7)3 = (25)3
6
3 6  6 C 1 .3 5 .21  6 C 2 .3 4 .2 2  6 C 3 .3 3 .2 3  6 C 4 3 2 .2 4  6 C 5 3 .2 5  6 C 6 2 6   3  2   5 6  25 

Nr (25) 3

1
Dr (25) 3
3
Ans.
1
JEEMAIN.GURU
JEE-Mathematics
Illustration 4 :
If in the expansion of (1 + x)m (1 – x)n, the coefficients of x and x2 are 3 and – 6 respectively then
m is -
[JEE 99]
(A) 6
Solution :
(B) 9
(C) 12
(D) 24


(m)(m  1).x 2
n(n  1) 2

 ...... 1  nx 
x  ......
(1 + x)m (1 – x)n = 1  mx 
2
2


 
Coefficient of x = m – n = 3
........(i)
n(n  1) m (m  1)

 6
2
2
Coefficient of x2 = –mn +
........(ii)
Solving (i) and (ii), we get
m = 12 and n = 9.
Do yourself - 1 :
(i)
 2 x
Expand  3x  
2

5
(ii) Expand (y + x)n
1
Pascal's triangle : A triangular arrangement of numbers as shown. The
1
numbers give the coefficients for the expansion of (x + y)n. The first row
1
is for n = 0, the second for n = 1, etc. Each row has 1 as its first and
1
last number. Other numbers are generated by adding the two numbers
IMPORTANT TERMS IN THE BINOMIAL EXPANSION
(a)
3
1
3
1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
e tc.
immediately to the left and right in the row above.
3.
1
2
:
General term: The general term or the ( r +1)th term in the expansion of (x + y)n is given by
Tr +1= nCr x n–r yr
11
 2 1
The coefficient of x 7 in the expansion of  ax  

bx 
(b)
1 

The coefficient of x –7 in the expansion of  ax  2 

bx 
11
Also, find the relation between a and b, so that these coefficients are equal.
Solution :
(a)
 2 1
In the expansion of  ax  
bx
Tr + 1 =
11
2 11–r
C r(ax )
 1
 
bx
11
, the general term is :
r
=
11
Cr.
a 11  r 22  3 r
.x
br
putting 22 – 3r = 7

3r = 15

T6 =
11
C5

r = 5
a6 7
.x
b5
 2 1
Hence the coefficient of x 7 in  ax  

bx 
11
is
Note that binomial coefficient of sixth term is
2
11
11
C 5a 6b –5.
C 5.
Ans.
Illustration 5 : Find : (a)
E
JEEMAIN.GURU
(b)
JEE-Mathematics
1 

In the expansion of  ax  2 

bx 
Tr + 1 =
11
C r(ax)
11–r
 1 
 2 
bx
11
, general term is :
r
= (–1) r 11 C r
a 11  r 11 3r
.x
br
putting 11 – 3r = –7

3r = 18

T 7 = (–1) 6.
r = 6

11
C6
a 5 7
.x
b6
1 

Hence the coefficient of x –7 in  ax  2 
bx
11
is
11
C 6a 5b –6.
Ans.
Also given :
 2 1
Coefficient of x in  ax  

bx 
11
7

11

ab = 1
= coefficient of x
–7
1 

in  ax  2 
bx
11
C 5 a 6 b –5 = 11 C 6 a 5 b –6
(
11
C 5 = 11 C 6 )
which is the required relation between a and b.
Ans.
Illustration 6 : Find the number of rational terms in the expansion of (9 1/4 + 8 1/6) 1000.
The general term in the expansion of (9 1/4 + 8 1/6) 1000 is
Solution :
Tr + 1 =
1000
Cr
1000  r
9 
1
4
8 
1
6
r
=
1000
Cr
3
1000  r
2
r
22
The above term will be rational if exponents of 3 and 2 are integers
It means
1000  r
r
and
must be integers
2
2
The possible set of values of r is {0, 2, 4, ..........., 1000}
Hence, number of rational terms is 501
E
(b)
Ans.
Middle term :
The middle term(s) in the expansion of (x + y)n is (are) :
(i)
If n is even, there is only one middle term which is given by T(n +2)/2= nCn/2. x n/2. yn/2
(ii)
If n is odd, there are two middle terms which are T(n +1)/2 &amp; T[(n+1)/2]+1
Important Note :
Middle term has greatest binomial coefficient and if there are 2 middle terms their coefficients will be equal.
When r= n if n is even
2

 nC r will be maximum
n–1 n+1
When r=
or
if n is odd
2
2

The term containing greatest binomial coefficient will be middle term in the expansion of (1 + x)n
3
JEEMAIN.GURU
JEE-Mathematics

x3 
Illustration 7 : Find the middle term in the expansion of  3x  

6

x3 
The number of terms in the expansion of  3x  

6
Solution :
 9 1 
i.e. 

 2 
th
T5 = T4

9 3 
and 

 2 
+1
9
is 10 (even). So there are two middle terms.
th
are two middle terms. They are given by T 5 and T 6
 x3 
= C 4 (3x)   
 6
9
9
4
5
=
9
x 12
9.8.7.6 3 5
189 17
.
x 17 =
x
=
4
1.2.3.4 2 4 .3 4
8
6
C 43 5x 5.
5
x 15 9.8.7.6 3 4 19
 x3 
21 19
. 5 5x = –
x
T 6 = T 5+1 = 9C 5(3x) 4    = – 9C 43 4.x 4. 5 =
 6
1.2.3.4 2 .3
16
6
and
(c)
Ans.
Term independent of x :
Term independent of x does not contain x ; Hence find the value of r for which the exponent of x is zero.
10
Illustration 8 :
Solution :
 x
 3 
  2
The term independent of x in 
 2x  
 3

5
(A) 1
(B)
12
General term in the expansion is
r
 x2  3 
Cr    2 
 3   2x 
10  r
2
3r
10
is 10
(C)
(D) none of these
C1
35 r
3r
20
 10  r 
2
3
2
which is not an integer. Therefore, there will be no constant term.
10
 10 C r x 2
.
10  r
2
For constant term,
Ans. (D)
Do yourself - 2 :
10
(i)
 2 1
Find the 7 term of  3x  
3

(ii)
 2 3 
Find the term independent of x in the expansion :  2x  3 
x 

th
25
7
(i i i )
Find the middle term in the expansion of :
(d)
Numerically greatest term :
Let numerically greatest term in the expansion of (a + b) n be T r+1.

| Tr 1 | Tr

 Tr 1  Tr 2
where T r+1 = n C ra n–rb r
Solving above inequalities we get
Case I : When
(b)
 2 1
 2x  x 


n 1
n 1
1  r 
a
a
1
1
b
b
n 1
a is an integer equal to m, then Tm and Tm+1 will be numerically greatest term.
1
b
n 1
a is not an integer and its integral part is m, then T m+1 will be the numerically
1
b
greatest term.
Case II : When
4
6
 2x 3 

(a) 

 3 2x 
E
JEEMAIN.GURU
JEE-Mathematics
Illustration 9 : Find numerically greatest term in the expansion of (3 – 5x) 11 when x =
Solution :
1
5
n 1
n 1
1  r 
a
a
Using 1 
1
b
b
11  1
11  1
1  r 
3
3
1
1
5 x
5 x
solving we get 2 &lt; r &lt; 3
 r = 2, 3
so, the greatest terms are T 2+1 and T 3+1 .

Greatest term (when r = 2)
T 3 = 11 C 2.3 9 (–5x) 2 = 55.3 9 = T 4
From above we say that the value of both greatest terms are equal.
Ans.
Illustration 10 : Given T 3 in the expansion of (1 – 3x) 6 has maximum numerical value. Find the range of 'x'.
n 1
n 1
1  r 
Solution :
Using
a
a
1
1
b
b
6 1
7
1  2 
1
1
1
1
3x
3x
Let |x| = t
21t
21t
1  2 
3t  1
3t  1
 4t 1
 1 1
 21t
 0  t   , 


3
 3t  1
 3t  1
 3 4
 

15 t  2  0  t   ,  1    2 ,  
 21t  2


 3t  1
3  15
 3t  1


Common solution
 2 1
t   ,  
 15 4 
2   2 1
 1
x   , 

,
15  15 4 
 4
Do yourself -3 :
(i)
Find the numerically greatest term in the expansion of (3 – 2x)9, when x = 1.
E
(ii)
4.
 1 2x 
In the expansion of  

2 3 
possible integral values of n.
n
when x = –
1
, it is known that 3rd term is the greatest term. Find the
2
PROPERTIES OF BINOMIAL COEFFICIENTS :
n
(1+x)n = C0 + C1x + C2x 2 + C3x 3 +.........+Cnx n   n C r r r ; n  N
....(i)
r 0
where C0,C 1,C 2,............Cn are called combinatorial (binomial) coefficients.
(a)
The sum of all the binomial coefficients is 2n.
Put x = 1, in (i) we get
n
C0 + C1 + C2 + .............+ Cn = 2n 

n
Cr  0
....(ii)
r 0
(b)
Put x=–1 in (i) we get
n
C0 – C1 + C2–C3............+ Cn = 0 
 ( 1)
r n
r 0
5
Cr  0
...(iii)
JEEMAIN.GURU
JEE-Mathematics
(c)
(d)
The sum of the binomial coefficients at odd position is equal to the sum of the binomial coefficients at
even position and each is equal to 2n–1.
From (ii) &amp; (iii),
C0 + C2 + C4 +............ = C1 + C3 + C5+....... = 2n–1
n
n
n+1
C r + C r–1= C r
n
(e)
n
Cr
n  r 1

r
C r 1
(f)
n
Cr 
n
r
(g)
n
Cr 
r  1 n 1
. C r 1
n 1
n 1
n n 1
.
r r 1
C r 1 
n 2
C r 2  ....... 
n(n  1)(n  2).......(n  r  1)
r(r  1)(r  2)..........1
Illustration 11 : Prove that : 25C 10 + 24C 10 +........+10C 10 = 26C 11
Solution :
LHS = 10C 10 + 11 C 10 + 12C 10 + ..............+ 25 C 10
11

C 11 + 11C 10 + 12C 10 + .......+ 25C 10
12

C 11 + 12 C 10 +........+ 25 C 10
13

C 11 + 13 C 10 +......... 25 C 10
and so on.  LHS = 26C11
Aliter :
LHS = coefficient of x10 in {(1 + x)10 + (1 + x)11 +............... (1+ x)25}
10
16

10 {1  x}  1 
(1

x)

in 
1  x 1 


coefficient of x

(1  x )26  (1  x )10 
coefficient of x10 in 
x

26
10
coefficient of x11 in (1  x )  (1  x )  =
26
C11 – 0 =
26
C11
Illustration 12 : Prove that :
C1 + 2C
(ii)
C0 
(i)
n
n
n
L.H.S. =  r. n C r   r. . n 1 C r 1
r 1
r 1 r
2
C1 C 2
C
2 n 1  1

 .........  n 
2
3
n 1
n 1
n
= n
r 1
n 1
C r 1  n.  n 1 C 0  n 1 C 1  .......  n 1 C n 1 


= n . 2n–1
Aliter : (Using method of differentiation)
(1 + x) n = nC0 + nC1x + nC2x2 + ....... + nCnxn
..........(A)
Differentiating (A), we get
n(1 + x)n – 1 = C1 + 2C2x + 3C3x2 + ....... + n.Cnxn – 1.
Put x = 1,
C 1  2C 2  3C 3  ........  n.C n  n.2 n 1
n
(ii)
Cr
1 n n 1 n

Cr
L.H.S. = 

n  1 r 0 r  1
r 0 r  1
1 n n 1
1
 n 1 C 1  n 1 C 2  .......  n 1 C n 1  = 1 2 n 1  1 
C r 1 
=

 n 1 

n  1 r 0
n 1 
6
Solution :
+ 3C3 + ........ + nCn = n . 2n–1
(i)
E
JEEMAIN.GURU
JEE-Mathematics
Aliter : (Using method of integration)
Integrating (A), we get
C x2 C x3
C x n 1
(1  x) n 1
 C  C 0 x  1  2  ........  n
n 1
2
3
n 1
1
Put x = 0, we get, C = –
n 1
n 1
C x2 C x3
C x n 1
(1  x)
1
 C 0 x  1  2  ........  n

n 1
2
3
n 1
Put x = 1, we get
(where C is a constant)
C1 C 2
C
2 n 1  1

 ....... n 
2
3
n 1
n 1
Put x = –1, we get
C
C
1
C 0  1  2  ....... 
2
3
n 1
C0 
n
Illustration 13 : If (1 + x)n =  n C r x r , then prove that C 12  2.C 22  3.C 23  .........  n.C 2n 
r 0
Solution :
(1 + x)n = C0 + C1x + C2x2 + C2x3 + ........ + Cn xn
(2n  1)!
((n  1)!)2
.........(i)
Differentiating both the sides, w.r.t. x, we get
n(1 + x)n–1 = C1 + 2C2x + 3C2x2 + ......... + n.Cnxn –1
.........(ii)
also, we have
(x + 1)n = C0xn + C1xn – 1 + C2xn –2 + ......... + Cn
.........(iii)
Multiplying (ii) &amp; (iii), we get
(C1 + 2C2x + 3C3x2 + ........ + Cnxn – 1)(C0xn + C1xn –1 + C2xn – 2 + ......... + Cn) = n(1 + x)2n – 1
Equating the coefficients of xn – 1, we get
C 12  2C 22  3C 23  .........  n.C 2n  n . 2n 1 C n 1 =
(2n  1)!
Ans.
((n  1)!)2
Illustration 14 : Prove that : C 0 – 3C 1 + 5C 2 – ........(–1) n(2n + 1)C n = 0
Solution :
T r = (–1) r(2r + 1) nC r = 2(–1) rr . nC r + (–1) r nC r
n
n
n
T r = 2  ( 1) r .r. .n 1 C r 1   ( 1) r
r
r 1
r 0
E
n
n
n
r 1
r 0
C r = 2  ( 1) r .n 1 C r 1   ( 1) r . n C r
= 2  n 1 C 0 n 1 C 1  .....   n C 0  n C 1  ....... = 0

 

Illustration 15 : Prove that ( 2nC 0) 2 – ( 2nC 1) 2 + ( 2nC 2) 2 – .... + (–1) n ( 2nC 2n) 2 = (–1) n.
Solution :
(1 – x) 2n = 2n C 0 – 2n C 1x + 2nC 2 x 2 – ....+(–1) n 2n C 2nx 2n
2n
Cn
...(i)
and (x + 1) 2n = 2n C 0 x 2n + 2n C 1 x 2n–1 + 2n C 2 x 2n–2 +...+ 2n C 2n
....(ii)
Multiplying (i) and (ii), we get
(x 2 –1) 2n = ( 2n C 0 – 2nC 1x + .... + (–1) n 2n C 2nx 2n) &times; ( 2n C 0x 2n + 2nC 1 x 2n–1 + .... + 2n C 2n ) .... (iii)
Now, coefficient of x 2n in R.H.S.
= ( 2nC 0) 2 – ( 2nC 1) 2 + ( 2nC 2) 2 – ...... + (–1) n ( 2nC 2n) 2

General term in L.H.S., T r+1 = 2nC r(x 2) 2n – r(–1) r
Putting 2(2n – r) = 2n

r = n

T n+1 = 2n C nx 2n (–1) n
Hence coeffiecient of x 2n in L.H.S. = (–1) n. 2nC n
But (iii) is an identity, therefore coefficient of x 2n in R.H.S. = coefficient of x 2n in L.H.S.

( 2nC 0) 2 – ( 2nC 1) 2 + ( 2nC 2) 2 – .... + (–1) n ( 2nC 2n) 2 = (–1) n. 2nC n
7
JEEMAIN.GURU
JEE-Mathematics
Illustration 16 : Prove that : nC 0. 2nC n – nC 1. 2n–2Cn n + nC 2. 2n–4Cn n + .... = 2 n
Solution :
L.H.S. = Coefficient of x n in [ nC 0(1 + x) 2n – nC 1(1 + x) 2n – 2 ......]
= Coefficient of x n in [(1 + x) 2 – 1] n
= Coefficient of x n in x n(x + 2) n = 2 n
Illustration 17 : If (1 + x) n = C 0 + C 1 x + C 2 x 2 + ..... + C n x n then show that the sum of the products of the
Ci's taken two at a time represented by :
  CiC j
2n !
2.n !n !
is equal to 2 2n – 1 –
0 i j  n
Solution :
Since (C 0 + C 1 + C 2 +.....+ C n–1 + C n)
=
2
C 20  C 12  C 22  .....  C 2n 1  C 2n  2(C 0 C 1  C 0 C 2  C 0 C 3  ...  C 0 C n + C 1 C 2 + C 1 C 3 +...
+ C 1 C n+ C 2 C 3 + C 2 C 4+...+C 2 C n +.....+C n–1 C n )
n 2
(2 ) =
2n
Cn + 2   C iC j
0  i j  n
2n 1

Hence   C i C j  2
0  i j  n
2n !
2.n !n !
Ans.
2
Illustration 18 : If (1 + x) n = C 0 + C 1x + C 2 x 2 +.....+ C nx n then prove that    C i  C j  = (n – 1) 2nC n + 2 2n
0  i j  n
Solution :
  C i  C j 
L.H.S
2
0  i j  n
= (C 0 + C 1 ) 2 + (C 0 + C 2 ) 2 +....+ (C 0 + C n ) 2 + (C 1 + C 2 ) 2 + (C 1 + C 3 ) 2 +....+ (C 1 + C n ) 2 +
(C 2 + C 3 ) 2 + (C 2 + C 4 ) 2 +... + (C 2 + C n) 2 +....+ (C n – 1 + C n) 2
2
2
2
2
= n(C 0  C 1  C 2  ....  C n )  2   C i C j
0  i j  n
2n ! 
 2 n 1


= n. 2n C n + 2. 2
2.n !n ! 

= n . 2n C n + 2 2n – 2n C n = (n – 1) .
{from Illustration 17}
2n
C n + 2 2n = R.H.S.
Do yourself - 4 :
n  n  n 
n 
         ........    =
0
1
2
     
n 
(A) 2 n – 1
(B) 2nCn
(i)
(D) 2 n+1
(a)
3C0 – 8C1 + 13C2 – 18C3 + .......... upto (n + 1) terms = 0, if n  2.
(b)
2C 0 + 2 2
(c)
C 20 
C
C1
C
C
3 n 1  1
 2 3 2  2 4 3  ......  2 n 1 n 
2
3
4
n 1
n 1
C 12 C 22
C2
(2n  1)!

 ......  n 
2
3
n  1 ((n  1)!)2
MULTINOMIAL THEOREM :
n
n
n r r
Using binomial theorem, we have (x + a)n =  C r x a , n  N
r 0
n
n!
x n r a r =  n ! x s a r , where s + r = n
= 
(n

r)!r
!
r 0
r  s  n r !s!
This result can be generalized in the following form.
(x1 + x2 + ...... + xk)n =
n!
x1r1 x 2r2 ......x rkk
r1  r2 ..... rk  n r1 !r2 !.....rk !

8
If (1 + x)n = C0 + C1x + C2x2 + .......... + Cnxn, n  N. Prove that
(ii)
5.
(C) 2n
E
JEEMAIN.GURU
JEE-Mathematics
n!
r1 r2 r3
rk
The general term in the above expansion r !r !r !.....r ! . x 1 x 2 x 3 ......x k
1 2 3
k
The number of terms in the above expansion is equal to the number of non-negative integral solution of the
equation r1 + r2 + ....... + rk = n because each solution of this equation gives a term in the above expansion.
n  k 1
The number of such solutions is
C k 1
Particular cases :
(i)
(x + y + z)n =
n! r s t
x y z
r
!s!
t!
r  s  t n

n + 3 –1
The above expansion has
(ii)
(x + y + z + u)n =
There are
n+4–1
C3 – 1 =
n + 2
C2 terms
n!
xp y q zr us
p
!
q
!r
!s!
p  q  r  s n
C4–1 =

n+3
C3
terms in the above expansion.
Illustration 19 : Find the coefficient of x2 y3z4w in the expansion of (x – y – z + w)10
Solution :
(x – y – z + w)10 =
n!
(x) p (  y) q (  z) r (w ) s
p  q  r  s 10 p ! q !r !s!

We want to get x2y3z4w this implies that p = 2, q = 3, r = 4, s = 1

Coefficient of x2y3z 4w is
10!
(–1)3(–1)4 = –12600
2! 3! 4 ! 1!
Ans.
Illustration 20 : Find the total number of terms in the expansion of (1 + x + y)10 and coefficient of x2y3.
Solution :
Total number of terms =
Coefficient of x2y3 =
10+3–1
C3 – 1 =
12
C2 = 66
10!
 2520
2!  3!  5 !
Ans.
Illustration 21 : Find the coefficient of x5 in the expansion of (2 – x + 3x2)6.
Solution :
E
The general term in the expansion of (2 – x + 3x2)6 =
6! r
2 (  x) s (3x 2 ) t , where r + s + t = 6.
r !s ! t !
6! r
2  ( 1) s  (3) t  x s 2 t
r !s ! t !
=
For the coefficient of x5, we must have s + 2t = 5.
But, r + s + t = 6,

s = 5 – 2t and r = 1 + t, where 0  r, s, t  6.
Now t = 0 
r = 1, s = 5.
t = 1 
r = 2, s = 3.
t=2
r = 3, s = 1.

Thus, there are three terms containing x5 and coefficient of x5
=
6!
6!
6!
 21  ( 1)5  3 0 
 2 2  ( 1)3  31 
 2 3  ( 1)1  3 2
1! 5 ! 0!
2 ! 3 ! 1!
3! 1! 2 !
= –12 – 720 – 4320 = –5052.
Ans.
9
JEEMAIN.GURU
JEE-Mathematics
2n
n 1
r
Illustration 22 : If (1+x+x 2)n   a r x , then prove that
(a) ar = a2n–r
r 0
Solution :
(a)
(b)
1
 a r  2 (3 n  a n )
r 0
We have
2n
n
1  x  x    a x
2
r
....(A)
r
r 0
Replace x by
1
x
n

2n
1 1 

1 
1  x  2    a r  x 
x 

 
r 0

x
 x 1
n
2n
  a x
2n r
r
r 0
2n
2n
r 0
r 0
 a r x r   a r x 2n r

(b)
2
r
{Using (A)}
Equating the coefficient of x2n–r on both sides, we get
a2n–r = ar for 0 &lt; r &lt; 2n.
Hence
ar = a 2n–r.
Putting x=1 in given series, then
a 0 + a 1 + a 2 + .........+ a 2n = (1+1+1) n
a0 + a 1 + a 2 + ..........+ a 2n = 3 n
....(1)
But ar = a2n–r for 0 &lt; r &lt; 2n

series (1) reduces to
2(a0 + a 1 +a 2 + ........+ a n–1) + a n = 3 n.

a0 + a 1 +a2 + .......+ an–1 =
1 n
(3 – an)
2
Do yourself - 5 :
(i)
APPLICATION OF BINOMIAL THEOREM :
Illustration 23 : If  6 6  14 
2 n 1
= [N] + F and F = N – [N]; where [.] denotes greatest integer function, then
NF is equal to
(A) 20 2n+1
Solution :
Since  6 6  14 
(B) an even integer
2 n 1
(D) 40 2n+1
= [N] + F
Let us assume that f =  6 6  14 
Now, [N] + F – f
(C) odd integer
=  6 6  14 
2 n 1
2 n 1
; where 0  f &lt; 1.
–  6 6  14 
2n 1
2n
= 2  2 n 1 C 1  6 6  (14)  2 n 1 C 3  6 6 
2 n 2
(14) 3  ....

[N] + F – f = even integer.
Now 0 &lt; F &lt; 1 and 0 &lt; f &lt; 1
so
–1 &lt; F – f &lt; 1 and F – f is an integer so it can only be zero
Thus NF =  6 6  14 
2 n 1
6
6  14 
10
2n 1
= 20 2n+1 .
Ans. (A,B)
6.
Find the coefficient of x2y5 in the expansion of (3 + 2x – y)10.
E
JEEMAIN.GURU
JEE-Mathematics
Illustration 24 : Find the last three digits in 1150.
Solution :
Expansion of (10 + 1) 50 =
=
50
50
50
50
C 010 50 +
49
50
50
C 110 49 + ..... + 50C 4810 2 +
50
C 4910 +
50
C 50
3
C 0 10  C 1 10  ......  C 47 10 + 49 &times; 25 &times; 100 + 500 + 1



1000 K

1000 K + 123001

Last 3 digits are 001.
Illustration 25 : Prove that 22225555 + 55552222 is divisible by 7.
When 2222 is divided by 7 it leaves a remainder 3. So adding &amp; subtracting 35555, we get :
Solution :
5555
5555
E  2222
 3 5555
 5555 2222


  3


E1
E2
For E1 : Now since 2222–3 = 2219 is divisible by 7, therefore E1 is divisible by 7
( xn – an is divisible by x –a)
For E2 : 5555 when devided by 7 leaves remainder 4. So adding and subtracting 42222, we get :
E2 = 35555 + 42222 + 55552222 – 42222
= (243)1111 + (16)1111 + (5555)2222 – 42222
Again (243)1111 + 16 1111 and (5555) 2222 – 4 2222 are divisible by 7
( xn + an is divisible by x + a when n is odd)
Hence 22225555 + 55552222 is divisible by 7.
Do yourself - 6 :
(i)
Prove that 525 – 325 is divisible by 2.
(ii)
Find the remainder when the number 9100 is divided by 8.
(i i i )
Find last three digits in 19100.
(iv)
Let R  (8  3 7 )20 and [.] denotes greatest integer function, then prove that :
(a) [R] is odd
(v)
7.
E
(b)
R  [R ]  1 
1
(8  3 7 )20
Find the digit at unit's place in the number 171995 + 111995 – 71995.
BINOMIAL THEOREM FOR NEGATIVE OR FR ACTIONAL INDICES :
If n Q, then (1 + x)n = 1 + nx +
n(n  1) 2 n(n  1)(n  2) 3
x +
x + .......  provided | x | &lt; 1.
2!
3!
Note :
(i)
When the index n is a positive integer the number of terms in the expansion of ( 1+ x)n is finite i.e. (n+1)
&amp; the coefficient of successive terms are : nC0, nC1, nC2, ....... nCn
(ii)
When the index is other than a positive integer such as negative integer or fraction, the number of terms
in the expansion of (1+ x)n is infinite and the symbol nCr cannot be used to denote the coefficient of the
general term.
(iii)
Following expansion should be remembered (|x| &lt; 1).
(a)
(1 + x)-1 =1 – x + x2 – x3 + x4 - .... 
(b)
(1 – x)–1 =1 + x + x2 + x3 + x4 + .... 
(c)
(1 + x)-2 =1 – 2x + 3x2 – 4x3 + .... 
(d)
(1 – x)–2 =1 + 2x + 3x2 + 4x3 + .... 
(e)
(1 + x)–3 = 1 – 3x + 6x2 – 10x3 + ..... +
11
( 1) r (r  1)(r  2) r
x  ........
2!
JEEMAIN.GURU
JEE-Mathematics
(r  1)(r  2) r
x  ........
2!
The expansions in ascending powers of x are only valid if x is ‘small’. If x is large i.e. | x |&gt;1 then we
(f)
(iv)
(1 – x)–3 = 1 + 3x + 6x2 + 10x3 + ..... +
may find it convenient to expand in powers of 1/x, which then will be small.
8.
APPROXIM ATIONS :
(1 + x)n = 1 + nx +
n(n  1) 2 n(n  1)(n  2) 3
x +
x .......
1.2.3
1.2
If x &lt; 1, the terms of the above expansion go on decreasing and if x be very small, a stage may be reached
when we may neglect the terms containing higher powers of x in the expansion. Thus, if x be so small that its
square and higher powers may be neglected then (1 + x)n = 1 + nx, approximately.
This is an approximate value of (1 + x)n
Illustration 26 : If x is so small such that its square and higher powers may be neglected then find the approximate
Solution :
(1  3x )1 / 2  (1  x) 5 / 3
 4  x 1 / 2
(1  3x )1 / 2  (1  x) 5 / 3
1/2
4  x 
3
5x
x 1 
1
19  
2
3
x  1 
=  2 
1/2
2
6 
x

2 1  

4
1
=
x

4
1 / 2
=
1
19  
x
x 1  
 2 


2
6
8
1
x 19 
x 19
41
x = 1  
x = 1 –
x
 2  
2
4 6 
8 12
24
Illustration 27 : The value of cube root of 1001 upto five decimal places is –
=
(A) 10.03333
Solution :
(B) 10.00333
(C) 10.00033
1 

(1001) 1/3 = (1000+1) 1/3 = 10  1 

1000 
1/3
Ans.
(D) none of these
1 1
1 / 3(1 / 3  1) 1



 .....
= 10 1  .
3 1000
2!
1000 2


= 10{1 + 0.0003333 – 0.00000011 + .....} = 10.00333
Illustration 28 : The sum of 1 +
(A)
Solution :
1 1.3 1.3.5


 .... is 4 4.8 4.8.12
(B)
2
Ans. (B)
1
(C)
2
Comparing with 1 + nx +
(D) 2 3/2
3
n(n  1) 2
x  ....
2!
nx = 1/4
....... (i)
2
and
n(n  1)x
1.3
=
2!
4.8
or
1 1
3

nx(nx  x)
3
=
   x  
4 4
16
2!
32

1 3
1
1
 3
  x    x    
4
4
4 4
2
(by (i))
........(ii)
putting the value of x in (i)
n (–1/2) = 1/4  n = –1/2

sum of series = (1 + x) n = (1 – 1/2) –1/2 = (1/2) –1/2 =
12
2
Ans. (A)
value of
E
JEEMAIN.GURU
9.
JEE-Mathematics
EXPONENTIAL SERIES :
(a)
e is an irrational number lying between 2.7 &amp; 2.8. Its value correct upto 10 places of decimal is
2.7182818284.
(b)
Logarithms to the base ‘e’ are known as the Napierian system, so named after Napier, their inventor.
They are also called Natural Logarithm.
1

x x2 x 3

 ....... ; where x may be any real or complex number &amp; e = Lim  1  
e = 1 
n


n
1! 2 ! 3 !
x
(c)
ax = 1 
(d)
e = 1
(e)
10.
x
x2 2
x3 3
na 
n a 
n a  ....... , where a &gt; 0
1!
2!
3!
1 1 1
   .......
1! 2 ! 3 !
LOGARITHMIC SERIES :
(a)
n (1 + x) = x 
x2 x3 x4


 ....... , where –1 &lt; x  1
2
3
4
(b)
n (1 - x) = - x 
x2 x 3 x 4


 ....... , where –1  x  1
2
3
4
Remember :
E
1 1 1
   .......   n 2
2 3 4
(i)
1
(iii)
n2 = 0.693
(ii)
elnx = x ; for all x &gt; 0
(iv)
n10 = 2.303
2
1.
(i)
5
 x
 x
C 0x (3x ) + C 1(3x )    + 5C 2(3x 2) 3   
 2
 2
2 5
5
2 4
3
2 2
(ii) nC0y n + nC 1y n–1.x + nC 2.y n–2.x 2 + ..........+ nCn.x n
70 8
x ;
3
(ii)
25 ! 15 10
2 3 ;
10! 5 !
(i i i ) (a) –20; (b) –560x5, 280x2
2 :
(i)
3.
4.
5.
6.
(i) 4th &amp; 5th i.e. 489888
( i i ) n = 4, 5, 6
(i) C
(i) –272160 or – 10C5 &times; 5C2 &times; 108
(ii) 1
(i i i ) 8 0 1
(v) 1
13
4
 x
 x
 x
+ C 3(3x )    + 5C 4(3x 2) 1    + 5C 5   
2
2




 2
5
5
n
JEEMAIN.GURU
JEE-Mathematics
EXERCISE - 01
SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER)
n
1.
x

If the coefficients of x7 &amp; x8 in the expansion of 2   are equal , then the value of n is 3

(A) 15
(B) 45
(C) 55
(D) 56
n
2.
1

The sum of the binomial coefficients of 2 x   is equal to 256 . The constant term in the expansion
x

is -
(A) 1120
3.
(B) 2110
(C) 1210
(D) none
The sum of the coefficients in the expansion of (1  2x + 5x2)n is ' a ' and the sum of the coefficients in the
expansion of (1 + x)2n is b . Then (B) a = b2
(A) a = b
4.
(C) a2 = b
(D) ab = 1
Given that the term of the expansion (x1/3  x1/2)15 which does not contain x is 5 m where m  N , then m is
equal to (A) 1100
5.
(B) 1010
(D) none
  1  4 x  1 7  1  4 x  1 7 

 
  is a polynomial in x of degree 2
2
4 x  1  




 

1
The expression
(A) 7
6.
(C) 1001
(B) 5
1/3
In the binomial (2
(C) 4
(D) 3
1/3 n
+3
) , if the ratio of the seventh term from the beginning of the expansion to the
seventh term from its end is 1/6 , then n is equal to (A) 6
(B) 9
(C) 12
(D) 15
11
7.
3
2 
The term independent of x in the product (4  x  7x )  x  
x

is -
(A) 7. 11 C 6
8.
(B) 36. 11C6
(C) 35. 11C5
(D) –12. 211
If ‘a’ be the sum of the odd terms &amp; ‘b’ be the sum of the even terms in the expansion of (1 + x)n , then
(1  x&sup2;)n is equal to (A) a&sup2;  b&sup2;
13
C5 . 208 . 105
13
(C) 611
C6 . 207 . 104

(C)
2 43
(B) 26
p 
If    0 for p &lt; q, where p, q  W, then
q 
(A) 2n
13.
(B)
Number of rational terms in the expansion of
(A) 25
12.
(B) 3 . 610
(D) none
The greatest terms of the expansion (2x + 5y)13 when x = 10, y = 2 is (A)
11.
(D) none
The sum of the coefficients of all the even powers of x in the expansion of (2x2  3x + 1)11 is (A) 2 . 610
10.
(C) b&sup2;  a&sup2;

13
C4 . 209 . 104
(D) none of these
100

is -
(C) 27
(D) 28
(C) 22n–1
(D)
(C) 13
(D) 14
n
  2r  
r 0 

(B) 2n–1
2n
Cn
x
 47  5  52  j   x 
   
    , then y =
 4  j 1  3   y 
(A) 11
(B) 12
14
9.
(B) a&sup2; + b&sup2;
E
JEEMAIN.GURU
14.
JEE-Mathematics
If n  N &amp; n is even , then
(A) 2n
15.
1
1
1
1


 ...... 
=
1. (n  1) ! 3 ! (n  3) ! 5 ! (n  5) !
(n  1) ! 1 !
(B)
2 n 1
n!
(C) 2n n !
(D) none of these
Let R  (5 5  11) 31  I  ƒ , where I is an integer and ƒ is the fractional part of R, then R &middot; ƒ is equal to (A) 231
(B) 3 31
(C) 262
(D) 1
10
16.
(A)
17.
25C
(B)
12
211
11
(B)
n
If a n 

r 0
1
, then
n
Cr
n

r 0
(A) ( n -1) an
19.
25C
(C)
15
C0
C
C
C
 1  2  ......  10 is equal to (here Cr =
1
2
3
11
(A)
18.
 10  15 

 is equal to r  0  r  14  r 

The value of
n
10
211  1
11
10
(D)
25 C
(D)
311  1
11
11
C r)
(C)
311
11
r
equals Cr
[JEE 98]
(B) n an
400
The last two digits of the number 3
(A) 81
25C
(C) n an /2
(D) none of these
(C) 29
(D) 01
are -
(B) 43
SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS)
20.
21.
22.
E
If the coefficients of three consecutive terms in the expansion of (1 + x)n are in the ratio of 1 : 7 : 42, then
n is divisible by (A) 9
(B) 5
(C) 3
(D) 11

1 
In the expansion of  3 4  4 
6

20
-
(A) the number of irrational terms = 19
(B) middle term is irrational
(C) the number of rational terms = 2
(D) 9th term is rational
If (1 + x + x 2 + x 3)100 = a 0 + a1x + a 2x 2 + ......... + a 300x300, then (A) a0 + a1 + a2 + a3 +.......+ a300 is divisible by 1024
(B) a0 + a 2 + a 4 +.......+ a 300 = a 1 + a 3 + ....... + a 299
(C) coefficients equidistant from beginning and end are equal
23.
(D) a1 = 100
The number 101100  1 is divisible by (A) 100
24.
25.
If
9 
(B) 1000
80

(C) 10000
(D) 100000
n
= I+f
where I , n are integers and 0 &lt; f &lt; 1 , then -
(A) I is an odd integer
(B) I is an even integer
(C) (I + f) (1  f) = 1
(D) 1  f =

1 
In the expansion of  x 2 / 3 


x
9 
80

n
30
, a term containing the power x13 -
(A) does not exist
(B) exists and the co-efficient is divisible by 29
(C) exists and the co-efficient is divisible by 63
(D) exists and the co-efficient is divisible by 65
15
JEEMAIN.GURU
JEE-Mathematics
The co-efficient of the middle term in the expansion of (1 + x)2n is (A)
1.3.5 .7......(2 n  1)
2n
n!
(B)
(C)
(n  1) (n  2) (n  3) .... (2n  1) (2n)
1.2 .3.......... (n  1) n
(D)
CHE CK Y OU R G R ASP
2n
Cn
2 .6 .10 .14 ...... (4n  6) (4n  2)
1.2.3 .4 .....(n  1) . n
KEY
EXERCISE-1
Que.
1
2
3
4
5
6
7
8
9
10
Ans.
C
A
A
C
D
B
B
A
B
C
Que.
11
12
13
14
15
16
17
18
19
20
B
C
D
B,D
Ans.
B
B
C
B
C
D
Que.
21
22
23
24
25
26
A, B, C
A ,C , D
B,C,D
A,B, C,D
Ans.
A,B, C,D A,B, C,D
16
26.
E
JEEMAIN.GURU
JEE-Mathematics
EXERCISE - 02
BRAIN TEASERS
SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS)
1.
The coefficient of xr (0  r  n  1) in the expression :
(x + 2)n1 + (x + 2)n2 . (x + 1) + (x + 2)n3 . (x + 1)&sup2; + ...... + (x + 1)n1 is (A)
2.
n
Cr (2r  1)
(B)
n
Cr (2nr  1)
(C)
n
Cr (2nr + 1)
(D)
(B) odd &amp; of the form 3n
(C) odd &amp; of the form (3n  1)
(D) odd &amp; of the form (3n + 1)
4
2 12
The co-efficient of x in the expansion of (1  x + 2x )
(A)
4.
Cr (2r + 1)
If (1 + x + x&sup2;)25 = a0 + a1x + a2x&sup2; + ..... + a50 . x50 then a0 + a2 + a4 + ..... + a50 is (A) even
3.
n
12
C3
(B)
22
13
C3
is (C)
n
14
C4
(D)
12
C3 + 3
13
C3 +
14
C4
2
Let (1 + x ) (1 + x) = A0 + A1 x + A2 x + ...... If A0, A1, A2 are in A.P. then the value of n is (A) 2
(B) 3
(C) 5
(D) 7
nr
5.
If

nk
Cr = xCy then -
k 1
(A) x = n + 1 ; y = r
(B) x = n ; y = r + 1
(C) x = n ; y = r
6.
(D) x = n + 1 ; y = r + 1
t
Co-efficient of  in the expansion of ( + p)
m1
+ ( + p)
m2
( + q) + ( + p)m  3 ( + q)2 +...... ( + q)m  1 where
 q and p  q is m
(A)
7.

Ct pt  qt

m
(B)
p q
p q

Ct pt  qt

m
(D)
p q

C t pm  t  q m  t
pq
(C) 2
(D)  2
(B) 5
The value r for which
(C) 6
 30r  15r    30r 1  151   .......   300  15r  is maximum is/are
(B) 22
th
(D) 8
(C) 23
3 x
term in the expansion of   
 2 3
-
(D) 24
n
10.
If the 6
11.
value(s) of n can be (A) 11
(B) 12
(C) 13
(D) 14
In the expansion of (1 + x)n (1 + y)n (1 + z)n , the sum of the co-efficients of the terms of degree ' r ' is (B) n C
3
(A) n C r
12.
when x = 3 is numerically greatest then the possible integral
(C)
r3
3n
Cr
(D) 3 . 2nCr
 35  10  45  r   x 
   
    , then x – y is equal to  6  r 0  5   y 
(A) 39
(B) 29
s
13.

Number of terms free from radical sign in the expansion of (1 + 31/3 + 71/7)10 is -
(A) 21
E
m
(C)
(B)  1
(A) 4
9.

The co-efficient of x401 in the expansion of (1 + x + x2 + ...... + x9) 1 , (x &lt; 1) is (A) 1
8.

C t pm  t  q m  t
The value of
(C) 52
(D) 40
(C) 3 n
(D) 3(3 n – 1)
n
  n C ss C r
is -
r 0 s 1
r s
(A) 3 n – 1
(B) 3 n + 1
17
JEEMAIN.GURU
JEE-Mathematics
14.
 log
 3
In the expansion of  x  3 .2

2
x3


11
-
(A) there appears a term with the power x2
(B) there does not appear a term with the power x2
(C) there appears a term with the power x 3
1
3
The sum of the series (1&sup2; + 1).1! + (2&sup2; + 1).2! + (3&sup2; + 1).3! + ..... + (n&sup2; + 1).n! is (A) (n + 1) . (n + 2)!
(B) n . (n + 1)!
(C) (n + 1) .(n + 1)!
(D) none of these
(D) the ratio of the co-efficient of x3 to that of x 3 is
15.
3n
17.
18.
Set of values of r for which,
(A) 4 elements
18
Cr  2 + 2 .
18
Cr  1 +
18
Cr 
(B) 5 elements
C13 contains -
(C) 7 elements
BRAIN TEASER S
20
(D) 10 elements
KEY
EXERCISE-2
Que.
1
2
3
4
5
6
7
8
9
10
Ans.
B
A
D
A,B
B
B
B
C
B,C
B,C,D
Que.
11
12
13
14
15
16
17
18
Ans.
C
D
A
B,C,D
B
D
A,C
C
18
16.
1 

The binomial expansion of  x k  2 k  , n  N contains a term independent of x 
x 
(A) only if k is an integer
(B) only if k is a natural number
(C) only if k is rational
(D) for any real k
n
2
n
Let n  N. If (1 + x) = a0 + a1x + a2x + ........+ anx and an–3, an–2, an–1 are in AP, then (A) a1, a2, a3 are in AP
(B) a1, a2, a3 are in HP
(C) n = 7
(D) n = 14
E
JEEMAIN.GURU
JEE-Mathematics
EXERCISE - 03
MISCELLANEOUS TYPE QUESTIONS
FILL IN THE BLANKS
1.
The greatest binomial coefficient in the expansion of (a + b)n is ________ given that the sum of all the
coefficients is equal to 4096.
2.
The number 71995 when divided by 100 leaves the remainder ________.
15
3.
4.
5.
 2 1
The term independent of x in the expansion of  x   is ________ .
x

p n
np
If (1 + x + x&sup2; + ..... + x ) = a0+ a1x + a2x&sup2; + ..... + anp x
then a1+ 2 a2+ 3a3 + .... + npanp = ________ .
If (1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + x3 + ...... + xn)  a0 + a1x + a2x2 + a3x3 + ...... + amxm
m
then

a r has the value equal to ________ .
r 0
6.
7.
 1
2
If the 6th term in the expansion of the binomial  8 / 3  x log10
x
(1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + ...... +
8

x  is 5600, then x = ________ .

x100) when written in the ascending power of
x then the highest exponent of x is ________ .
MATCH THE COLUMN
Following question contains statements given in two columns, which have to be matched. The statements in
Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given
statement in Column-I can have correct matching with ONE statement in Column-II.
1.
Column-I
Column-II
(A)
(2n + 1) (2n + 3) (2n + 5) ....... (4n  1) is equal to
(p)
(n  1) n
n!
(B)
C1
2 . C2
3 . C3
n . Cn
is equal to


 ...... 
C0
C1
C2
C n 1
(q)
n . 2n . (2n  1)
(r)
(4n) ! n !
2 . (2n) ! (2n) !
(s)
n (n  1)
2
here Cr stand for
(C)
n
Cr .
If (C0 + C1) (C1 + C2) (C2 + C3) ...... (Cn1 + Cn)
n
= m . C1C2C3 .... Cn1 , then m is equal to
(D)
If Cr are the binomial coefficients in the expansion of
n
E
(1 + x)n, the value of
n
  (i  j) C C
i
j
is
i 1 j 1
ASSERTION
&amp;
REASON
These questions contains, Statement I (assertion) and Statement II (reason).
(A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
(B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I
(C) Statement-I is true, Statement-II is false
(D) Statement-I is false, Statement-II is true
1.
Statement-I : Coefficient of ab 8 c 3 d 2 in the expansion of (a + b + c + d) 14 is 180180
Because
Statement-II : General term in the expansion of (a 1 + a 2 + a 3 + ..... + a m ) n
n!
n n
a 1 a 2 ...a nmm , where n 1 + n 2 + n 3 + ... + n m = n.
= 
n1 !n 2 !n 3 !....n m ! 1 2
(A) A
(B) B
(C) C
19
(D) D
JEEMAIN.GURU
JEE-Mathematics
2.
Statement-I : If q =
1
and p + q = 1, then
3
15
r
15
C r p r q 15  r  15 
r 0
1
5
3
Because
n
n
r
Statement-II : If p + q = 1 , 0 &lt; p &lt; 1, then
C r p r q n r  np
r 0
(A) A
3.
(B) B
(C) C
(D) D
Statement-I : The greatest value of 40 C 0 . 60 C r + 40 C 1 . 60 C r–1 ........ 40 C 40 . 60 C r–40 is 100 C 50
Because
Statement-II : The greatest value of 2n C r , (where r is constant) occurs at r = n.
(A) A
(B) B
(C) C
(D) D
4.
x 1
Statement-I : If x = n C n–1 + n+1 C n–1 + n+2 C n–1 + .......... + 2n C n–1 , then
is integer..
2n  1
Because
Statement-II : n C r + n C r–1 = n+1 C r and n C r is divisible by n if n and r are co-prime.
(A) A
(B) B
(C) C
(D) D
COMPREHENSION BASED QUESTIONS
Comprehension # 1
2n
If n is positive integer and if (1 + 4x + 4x 2) n =
 ar xr ,
where a i's are (i = 0, 1, 2, 3, ..... , 2n) real
r 0
numbers.
On the basis of above information, answer the following questions :
n
1.
The value of 2
 a 2r is
-
r 0
(A) 9 n – 1
(B) 9 n + 1
(C) 9 n – 2
(D) 9 n + 2
n
2.
The value of 2
 a 2r 1
is -
3.
4.
(A) 9 n – 1
(B) 9 n + 1
(C) 9 n – 2
(D) 9 n + 2
The value of a 2n–1 is (A) 2 2n
(B) (n – 1).2 2n
(C) n.2 2n
(D) (n + 1).2 2n
(B) 8n 2 – 4
(C) 8n 2 – 4n
(D) 8n – 4
The value of a 2 is (A) 8n
M ISCEL L AN E OU S TYP E Q U ESTION

12
C6
2. 43
3 . 3003
4.
np
(p + 1) n
2
5 . (n + 1) !
Matc h th e C o lu mn
1 . (A)(r), (B)(s), (C)(p), (D)(q)

A s s er ti o n &amp; R eas o n
1. C

EXERCISE-3
F i ll i n t h e B lanks
1.

KEY
2. D
C o mp r eh e ns i o n
3. C
B as ed
Comprehensi on # 1 :
4. A
Qu es ti o ns
1. B
2. A
3. C
20
4. C
6 . x = 10
7 . 5050
r 1
E
JEEMAIN.GURU
JEE-Mathematics
EXERCISE - 04 [A]
CONCEPTUAL SUBJECTIVE EXERCISE
1.
If the coefficients of (2r + 4)th, (r - 2)th terms in the expansion of (1 + x)18 are equal, find r.
2.
If the coefficients of the rth, (r +1)th &amp; (r + 2 )th terms in the expansion of (1 + x)14 are in AP, find r.
3.
 x
3 
Find the term independent of x in the expansion of : (a) 
 2
 3 2x 
4.
Prove that :
5.
If
40
6.
If
n + 1
7.
Which is larger : (9950 +10050) or (101)50.
8.
Show that
9.
Find the coefficient of x4 in the expansion of :
10
(a)
10.
n -1
n -2
Cr +
Cr+
C1 . x(1  x)39 + 2 .
40
n -3
Cr +....... + rCr=
C2 x2 (1  x)38 + 3
40
n
1 1 / 3

 x 1 / 5 
(b)  x
2

8
Cr+1.
C3 x3 (1  x)37 + ...... + 40.
40
C40 x40 = ax + b, then find a &amp; b.
C2 + 2 (2C2 + 3C2 + 4C2 + ...... + nC2) = 12 + 22 + 32 + ......... + 1002, then find n.
2n -2
Cn-2 + 2.
2n -2
Cn-1+
2n -2
Cn &gt;
4n
, n N, n &gt;2
n 1
(1 + x + x2 + x3)11
(b)
(2 – x + 3x2)6
Find numerically the greatest term in the expansion of :
(a)
( 2 + 3x)9 when x =
3
2
(b)
( 3 – 5x)15 when x =
1
5
11.
2

Prove that the ratio of the coefficient of x10 in (1– x2)10 &amp; the term independent of x in  x  

x
12.
 3x 2
1 
  .
Find the term independent of x in the expansion of (1+ x +2x ) 
3x 
 2
13.
Prove that
10
is 1 : 32.
9
3
n

n
C k sin Kx. cos ( n - K)x = 2n -1 sin nx.
k 0
14.
E
Find the coefficient of :
(a)
x6 in the expansion of (ax2 + bx + c)9.
(c)
a2 b3 c4d in the expansion of (a – b – c + d)10.
(b)
x2 y3 z4 in the expansion of (ax - by + cz)9.
25
20
 20  30 
  r  25  r   x C y , then find x, y. (b)


r 0 
15.
(a)
16.
 n  r   n  n  k 

Prove that :      
 r  k   k  r  k 
17.
Prove that :
25
  1 
r 0
r
Prove that :
 30  30 
 
0
 r  25  r 
n 2
18.
Prove that :
 n  1  n   2n  1 



 n 2 
r 0  r  r  2 

21
 30   70  100

  C 25
r 0  r   25  r 

JEEMAIN.GURU
JEE-Mathematics
Prove the following (here C r = n C r) (Q. 19 to 26) :
(2n)!
(n  1)!(n  1)!
19.
C 0C 1 + C 1C 2 + C 2C 3 +.......+ C n-1C n =
20.
C 0 C r  C 1 C r 1  C 2 C r 2  .......  C n  r C n 
21.
C02 + C 12 + C 22 +....... + C n2 =
22.
C 0  C 1  C 2  C 3  .......  ( 1) r .C r 
23.
C1 + 2C2+ 3C3 + ....... + n. Cn =n. 2n-1
24.
C0 + 2C1+ 3C2 +....... + (n +1) Cn= (n + 2) 2n-1
25.
C0 
26.
C 1 2C 2 3C 3
n.C n n(n  1)


 ...... 

C0
C1
C2
C n 1
2
27.
Prove the identity
28.
If (1 + x )15 = C0 + C1. x + C2. x2 +....+ C15. x15 and C2 + 2C3 + 3C4 +....+ 14C15 = a2 + c, then find a + b + c.
29.
 30   15 
15  30  30 
14  30  29 
13  30  28 
Evaluate : 2     2     2    ......     
 0  15 
 1  14 
 2  13 
 15   0 
2n !
(n  r)!(n  r)!
(2n)!
n !n !
( 1) r (n  1)!
r !.(n  r  1)!
C1 C 2
C
2 n 1  1

 .......  n 
2
3
n 1
n 1
1
2 n 1
Cr

1
2n  2

C r 1 2n  1
2 n 1
1
.
Cr
2n
CON CEP TUAL SU BJ ECTIVE E X ER CISE
1. r = 6
2.
r = 5 or 9
7 . 101 5 0
9.
(a) 990
3.
(a) T3 =
x = 50, y = 25
(b) T6 = 7
28. 28
EXERCISE-4(A)
5. a = 40, b = 0
17
7.3 13
(b) 455 x 312
12.
54
2
(b) –1260.a2b 3c4 ; (c) –12600
(b) 3660 1 0 . (a) T7 
1 4 . (a) 84b 6c 3 + 630ab 4c 4 +756a 2 b 2c 5 + 84a 3c 6 ;
1 5 (a)
5
12
KEY
 30 
29.  
 15 
22
6. 100
b
E
JEEMAIN.GURU
JEE-Mathematics
EXERCISE - 04 [B]
2n
1.
If
a
r 0
2.
3.
BRAIN STORMING
SUBJECTIVE EXERCISE
2n
r
(x  2) r  b r (x  3) r &amp; a =1 for all k  n, then show that b =
k
n
r 0
2n+1
Cn+1
Prove the following :
(a)
if n is odd
0
2
2
2
2
2
C 0  C 1  C 2  C 3  .......  ( 1) n C n  
n/2 n
C n / 2 if n is even
( 1)
(b)
1.C 02 + 3.C 12 + 5.C 22 +.......+(2n+1)C n2 =
(a)
 x 2
Find the index n of the binomial   
5 5
(n  1)(2n)!
n !n !
n
if the 9th term of the expansion has numerically the greatest
coefficient (n N).
(b)
4.
For which positive values of x is the fourth term in the expansion of (5 +3x)10 is the greatest.
If a0, a1, a2, ....... be the coefficients in the expansion of (1 + x + x2)n in ascending powers of x, then prove that :
(a)
a0a1 – a1a 2 + a2a3 – .... = 0
(b)
a0a2 – a 1a 3+ a 2a 4 – ....... + a 2n-2 a 2n = a n+1 or a n-1
(c)
E1 = E 2 = E 3 = 3 n-1 ; where E 1 = a 0+ a 3+a 6+ ....... ; E 2 = a 1 + a 4+ a 7 +....... &amp; E 3 = a 2+ a 5+a 8+.......
5.
Prove that : 12 . C0 + 22 C1 + 32. C2 + 42. C3. +........+(n +1)2 . Cn= 2n-2 ( n +1) (n + 4).
6.
If (1+ x)n =
n
 C .x
r
r
r 0
2
3
4
n 2
3 n  2  2n  5
then prove that ; 2 .C 0  2 .C 1  2 .C 2  .......  2 .C n

1.2
2.3
3.4
(n  1)(n  2) (n  1)(n  2)
r
n  i n  r 1  n 



 k 1  k 1
i 0  k 
7.
Prove that :

8.
Prove that :
  j  n  j    p  n  , p, q  N;

 p  q 
j 0 
n
E


 n  1  n 
p  q 

p, q are constants.

 2n  1 
  n  r  r    n  1 
9.
Prove that :
10.
Prove that :
C 0 C1 C 2 C 3
C
1  n. 2 n 1



 .......  n 
2
3
4
5
n  2 (n  1)(n  2)
11.
Prove that :
1
2
12.
Prove that : (2nC 1)2 +2. (2nC 2)2 +3.( 2nC3)2 +..... +2n.(2nC 2n) 2 =
r 1 
n
 
C1 
2
3
n

C2 

3
4
n
C3 
BRAIN STOR MIN G SUBJ ECTIVE E X ER CISE
3.
(a)
n = 12
(b)
4
5
n
C 4  ..... 
5
20
x
8
21
23
( 1) n 1 n n
1
. Cn 
n 1
n 1
(4n  1)!
[(2n  1)!]2
KEY
EXERCISE-4(B)
JEEMAIN.GURU
JEE-Mathematics
EXERCISE - 05 [A]
The sum of the coefficients in the expansion of (x + y)n is 4096. The greatest coefficient in the expansion
is[AIEEE 2002]
(1) 1024
2.
(2) 924
(2) n = 3r
n
n n  1
(2)
2
2

n n  1
2
3 85
256

 n  1  n  2 
(4)
2
is-
[AIEEE
(3) 34
5
3
(2)
10
3
(3)
3
10
3
5
(4)
(2) (–1) n (1–n)
[AIEEE
(3) (–1) n–1 (n –1) 2
If the coefficients of r th, (r + 1) th and (r + 2) th terms in the binomial expansion (1 + y) m are in A.P., then
m and r satisfy the equation[AIEEE 2005]
(1) m 2 – m (4r – 1) + 4r 2 + 2 = 0
(2)
(3) m 2 – m (4r + 1) + 4r 2 + 2 = 0
(4) m 2 – m(4r – 1) + 4r 2 – 2 = 0

 1 
If the coefficient of x7 in ax 2    
 bx  

m 2 – m (4r + 1) + 4r 2 – 2 = 0

 1
equals the coefficient of x–7 in ax  
 bx 2

11



, then a and b satisfy
[AIEEE
(2)
(1) (45, 35)
The sum of the series
(1)
11.
2005]
a
1
(3) a + b = 1
(4) a – b = 1
b
For natural numbers m, n if (1 – y) m (1 + y) n = 1 + a 1 y + a 2 y 2 +......, and a 1 = a 2 = 10, then (m, n)
[AIEEE 2006]
is-
(1) ab = 1
10.
2004]
(4) (–1) n–1 n
the relation-
9.
2003]
(4) 35
The coefficient of x n in expansion of (1 + x)(1 – x) n is-
11
8.
(3)
(2) 33
(1) (n – 1)
7.
n n  2 
[AIEEE-2002]
The coefficient of the middle term in the binomial expansion in powers of x of (1 + x) 4 and of
(1 – x) 6 is the same if  equals[AIEEE 2004]
(1) 
6.
(4) n = 2r – 1
C1 2C 2 3C 3
nC n


 ... 

C0
C1
C2
C n 1
The number of integral terms in the expansion of
(1) 32
5.
(3) n = 2r + 1
If 1  x   C 0  C 1 x  C 2 x 2  ...  ...C n x n , then
(1)
4.
(4) 724
If for positive integers r &gt; 1, n &gt; 2 the coefficients of the (3r) th and (r+2) th powers of x in the expansion
[AIEEE 2002]
of (1+x) 2n are equal, then(1) n = 2r
3.
(3) 824
1 20
C 10
2
(2) (35, 45)
20 C
0

20 C
1
+
(3) (20, 45)
20 C
2

20 C
3
+ ..... ...... +
(3) – 20 C 10
(2) 0
(4) (35, 20)
20 C
10
is -
[AIEEE 2007]
(4) 20 C 10
In the binomial expansion of (a  b) n , n  5, the sum of 5 th and 6 th terms is zero, then
a
equals
b
[AIEEE
(1)
6
n 5
(2)
n 5
6
(3)
24
n4
5
(4)
5
n4
2007]
1.
JEE-[MAIN] : PREVIOUS YEAR QUESTIONS
E
JEEMAIN.GURU
JEE-Mathematics
n
12.
 r  1
Statement –1 :
n
C r   n  2  2 n 1
r 0
n
Statement–2 :
  r  1
n
C r x r = (1 + x)n+nx (1+x) n–1
[AIEEE 2008]
r 0
(1) Statement –1 is false, Statement –2 is true
(2) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1
(3) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement–1
(4) Statement–1 is true, Statement–2 is false
13.
The remainder left out when 8 2n – (62) 2n+1 is divided by 9 is :(1) 7
(2) 8
10
14.
[AIEEE 2009]
(3) 0
10
10
(4) 2
10
Let S 1   j( j  1)10 C j , S 2   j10 C j and S 3   j2 C j .
j 1
j 1
[AIEEE-2010]
j 1
Statement–1 : S 3 = 55 &times; 2 9 .
Statement–2 : S 1 = 90 &times; 2 8 and S 2 = 10 &times; 2 8 .
(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.
15.
The coefficient of x 7 in the expansion of (1 – x – x 2 + x 3)6 is :(1) –144
16.
(2) 132
If n is a positive integer, then

[AIEEE 2011]
(3) 144
2n
3  1
2n
  3  1
(4) – 132
is :
[AIEEE 2012]
(1) a rational number other than positive integers
(2) an irrational number
(3) an odd positive integer
(4) an even positive integer
10
17.
E

x 1
x 1 
The term independent of x in expansion of 


2/3
1/3
x
x
 1 x  x1 / 2 
(1) 4
(2) 120
(3) 210
P RE VIOU S Y EARS QU E STION S
Q ue.
1
is :
[JEE
(Main)-2013]
(4) 310
KEY
E XE R CISE -5
[A]
2
3
4
5
6
7
8
9
10
11
12
13
14
15
3
2
3
2
2
1
2
1
3
2
4
3
1
Ans
2
3
Q ue.
16
17
Ans
2
3
25
JEEMAIN.GURU
JEE-Mathematics
EXERCISE - 05 [B]
1.
 n
 n 
 n 
( a ) For 2 rn ,   + 2 
+ 
=

r 
 r  1
 r  2 
 n  1
(A) 
r  1 
[JEE 2000, (Screening ), 1+1M]
 n  1
(B) 2 
r  1 
 n  2
(C) 2 
r 
 n  2
(D) 
r 
( b ) In the binomial expansion of (a - b)n, n  5, the sum of the 5th and 6th terms is zero, Then
n5
6
(A)
2.
(B)
n4
5
5
n4
(C)
 n 
For any positive integers m , n (with n  m) , let   =
m
(D)
 n 2
+ 3 
m 
Cm . Prove that :
3.
The sum
m  n 2
+ ........ + (n  m + 1)   = 
.
m
m  2 
 10   20 
p
(where
 0 ) if p &lt; q is maximum when m is i   m  i 
i 0
q
(A) 5
4.
(B) 10
( a ) Coefficient of t
(A)
12
[JEE 2000 (Mains), 6M]
FG IJ
HK
m
 
24
in the expansion of (1 + t )
C6 + 2
(B)
[JEE 02(Screening ), 3M]
(C) 15
2 12
12
6
n5
n
 n 
 n 1
 n 2
 n 1 
m
 m  +  m  +  m  + ........ +   =  m  1 
m
Hence or otherwise prove that,
 n 
 n 1
 m  + 2  m 
a
equals b
12
(D) 20
24
(1 + t ) (1 + t ) is -
C6 + 1
12
(C)
[JEE 03, Screening , 3M out of 60]
C6
(D) none
( b ) If n and k are positive integers, show that
[JEE 03, Mains 2M out of 60]
 n  n
 n   n 1 
 n   n 2 
 n   n k   n 
2 k     2 k 1      2 k 2     .....( 1) k       
 0  k
 1   k 1 
 2   k 2 
 k  0   k
5.
If n, r  N and
n–1
Cr = (k2 – 3) (nCr+1), then k lies in the interval [JEE 04, Screeni ng, 3M out of 84]
6.
3 
The value of
(C)   3,  
(B) (2,  )
(D)

3, 2 

FG30IJ FG 30IJ  FG 30IJ FG 30 IJ  FG 30IJ FG 30 IJ ..........FG 30 IJ FG30 IJ , is where FG nIJ 
Hr K
H 0 K H10 K H 1 K H 11K H 2 K H 12K
H 20K H 30K
n
Cr
[JEE 05, Screeni ng, 3M out of 84]
30 C
7.
31 C
(A)
(B)
(C)
(D) 30C11
10
20
11 or
10
r
10
For r = 0, 1,....,10, let Ar, Br and Cr denote, respectively, the coefficient of x in the expansions of (1+x) ,
(1 + x)
20
60 C
30
31 C
10
and (1 + x) . Then
A
r
(B 10 B r  C 10 A r ) is equal to -
[JEE 10, 5M, –2M]
r 1

2
(B) A 10 B 10  C 10 A 10
(A) B 10 – C 10
8.

(C) 0
The coefficients of three consecutive terms of (1 + x)
(D) C 10 – B 10
n+5
are in the ratio 5 : 10 : 14. Then n =
P RE VIOU S Y EARS QU E STION S
1. (a) D ; (b) B
3. C
4. (a) A
5. D
26
KEY
6. A
E XE R CISE -5
7.
D
8.
6
[B]