binomial theorem-jee(main)
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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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 3
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 5
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 7
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.
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 9
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 11
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 13
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 15
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 17
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 19
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 21
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 23
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.
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 25
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 27
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 29
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 31
= !)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 ]
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 33
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
RESONANCE JEE(MAIN) BINOMIAL THEOREM - 35
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