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BINOMIAL THEOREM
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Nature of Chapter:
1.
You will find this chapter very interesting in the beginning.
Make sure you get good hold of the first theorem ie Binomial
theorem.
2.
This chapter is all about exploring different applications of
Binomial theorem.
3.
It will involve a lot of calculations (although standard ones).
Lengthy calculations are seen in Jee.
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Weightage of Sequence and series (Last 5 years)
2023
2022
2021
2020
2019
Average
JEE Main
6.4 %
4.2 %
5.0 %
4.8 %
4.6 %
5.00 %
Jee Advanced
3%
0%
0%
3%
2%
1.60 %
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Binomial Theorem
●
Binomial theorem for positive integral index
●
General term of Binomial expansion
●
Application of Binomial theorem
●
Greatest term in expansion of (a + bx)n
●
R-f Factor Theorem
●
Binomial for any index
●
Multinomial theorem
●
Binomial coefficient problems
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Binomial Theorem
Critical Topics in the Chapter
●
General term of Binomial expansion
●
Remainder problems
●
Binomial Coefficient Problems
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Notations: nCr
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Notations: nCr
E.g:
(a)
(b)
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Notations: nCr
E.g:
(a)
(c)
(b)
(d)
Clearly,
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Notations: nCr
Result
(a)
(b)
(c)
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Notations: nCr
(a)
Observation
(a)
(b)
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Notations: nCr
(b)
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Notations: nCr
(b)
Observation
(a)
(b)
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Q
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Q
Solution:
Given, 2nC5 = 2nC7
⇒ 5 + 7 = 2n
n=6
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JEE Main 6th Apr, 2023
Q
If 2nC3 : nC3 = 10 : 1, then the ratio
(n2 + 3n) : (n2 - 3n + 4) is:
A
27 : 11
B
35 : 16
C
2:1
D
65 : 37
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JEE Main 6th Apr, 2023
Q
If 2nC3 : nC3 = 10 : 1, then the ratio
(n2 + 3n) : (n2 - 3n + 4) is:
A
27 : 11
B
35 : 16
C
2:1
D
65 : 37
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JEE Main 6th Apr, 2023
Q
If 2nC3 : nC3 = 10 : 1, then the ratio
(n2 + 3n) : (n2 - 3n + 4) is:
Solution:
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Q
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Q
Solution:
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Binomial Theorem for Positive integral Index
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Binomial Theorem for Positive integral Index
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Binomial Theorem for Positive integral Index
=
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Binomial Theorem for Positive integral Index
=
Observations
(a) There are n + 1 terms in the expansion
(b) Sum of powers of x & y in each term is n.
(c) Coefficients of the terms equidistant from beginning
& end are equal, as nCr = nCn - r
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Binomial Theorem for Positive integral Index
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Binomial Theorem for Positive integral Index
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Binomial Theorem for Positive integral Index
Also, try to observe:
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Binomial Theorem for Positive integral Index
Also, try to observe:
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Q
Use binomial theorem to evaluate:
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Q
Use binomial theorem to evaluate:
Solution:
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General Term of Binomial Expansion
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General Term of Binomial Expansion
General term,
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General Term of Binomial Expansion
General term,
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Q
If the 21st and 22nd terms in the expansion of (1 - x)44 are equal,
then find the value of x
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Q
If the 21st and 22nd terms in the expansion of (1 - x)44 are equal,
then find the value of x
Solution:
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JEE Main 2019
Q
If the third term in the binomial expansion of
equals 2560, then a possible value of x is
A
B
C
D
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JEE Main 2019
Q
If the third term in the binomial expansion of
equals 2560, then a possible value of x is
A
B
C
D
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JEE Main 2019
Q
If the third term in the binomial expansion of
equals 2560, then a possible value of x is
Solution:
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Q
Find the coefficient of x7 in the expansion of
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Q
Find the coefficient of x7 in the expansion of
Solution:
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Q
What are the values of k if the term independent of x in
the expansion of
is 405?
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Q
What are the values of k if the term independent of x in
the expansion of
Solution:
is 405?
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Q
The coefficient of x53 in the expansion
A
100C
B
100C
C
-100C53
D
-100C100
47
53
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Q
The coefficient of x53 in the expansion
A
100C
B
100C
C
-100C53
D
-100C100
47
53
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Q
The coefficient of x53 in the expansion
Solution:
The given sigma is the expansion of
[(x - 3) + 2]100 = (x - 1)100 = (1 - x)100
Therefore, x53 will occur in T54.
T54 = 100C53(-x)53
Therefore, the coefficient is - 100C53
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Q
JEE Advanced 2023, P1
Let a and b be two non-zero real numbers. If the coefficient of x5 in
the expansion of
is equal to the coefficient of x-5
in the expansion of
, then the value of 2b is
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Q
JEE Advanced 2023, P1
Let a and b be two non-zero real numbers. If the coefficient of x5 in
the expansion of
is equal to the coefficient of x-5
in the expansion of
, then the value of 2b is
Ans: 3
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Solution:
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Q
Find number of terms in the expansion of
which are integers.
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Q
Find number of terms in the expansion of
which are integers.
Ans: 32
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Q
Find number of terms in the expansion of
which are integers.
Solution:
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Q
Find the 4th term from the end in the expansion of
Remark
kth term from the end in expansion of (x + y)n is (n - k + 2)th term
from beginning
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Q
Find the 4th term from the end in the expansion of
Solution:
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Q
Find the 4th term from the end in the expansion of
Alternate Solution:
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Middle Term of Binomial Expansion
Now, lets analyse middle term in expansion of (x + y)n
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Q
Find middle term(s) in the expansion of
(a)(x + y)6
(b) (x + y)13
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Q
Find middle term(s) in the expansion of
(a)(x + y)6
Solution:
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Q
Find middle term(s) in the expansion of
(b)(x + y)13
Solution:
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Middle Term of Binomial Expansion
Observation
Middle term of (1 + x)n is
when n is even
, when n is odd
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JEE Main 17th Mar, 2021
Q
If the middle term in the expansion of
is equal to
, then x is equal to; (n ∈ I)
A
B
C
D
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JEE Main 17th Mar, 2021
Q
If the middle term in the expansion of
is equal to
, then x is equal to; (n ∈ I)
A
B
C
D
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JEE Main 17th Mar, 2021
Q
If the middle term in the expansion of
is equal to
Solution:
, then x is equal to; (n ∈ I)
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General Term of Binomial Expansion
There are two particular cases which are used very frequently.
Lets see those cases.
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General Term of Binomial Expansion
There are two particular cases which are used very frequently.
Lets see those cases.
(a)
(b)
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Q
If in the expansion of (1 + x)m (1 - x)n the coefficient
of x is 3, then
A
m-n=3
B
m+n=3
C
m-n=9
D
mn + m + n = 3
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Q
If in the expansion of (1 + x)m (1 - x)n the coefficient
of x is 3, then
A
m-n=3
B
m+n=3
C
m-n=9
D
mn + m + n = 3
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Q
If in the expansion of (1 + x)m (1 - x)n the coefficient
of x is 3, then
Solution:
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Q
The coefficient of x18 in the product (1 + x) (1 - x)10(1 + x + x2)9 is
A
84
B
-126
C
-84
D
126
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Q
The coefficient of x18 in the product (1 + x) (1 - x)10(1 + x + x2)9 is
A
84
B
-126
C
-84
D
126
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Q
The coefficient of x18 in the product (1 + x) (1 - x)10(1 + x + x2)9 is
Solution:
Thus, the coefficient of x18 in the product is
equal to
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Single Correct Question
Q
JEE Advanced 2014, P2
Coefficient of x11 in the expansion of (1 + x2)4(1 + x3)7(1 + x4)12 is
A
1051
B
1106
C
1113
D
1120
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Single Correct Question
Q
JEE Advanced 2014, P2
Coefficient of x11 in the expansion of (1 + x2)4(1 + x3)7(1 + x4)12 is
A
1051
B
1106
C
1113
D
1120
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JEE Advanced 2014, P2
Single Correct Question
Q
Coefficient of x11 in the expansion of (1 + x2)4(1 + x3)7(1 + x4)12 is
Solution:
Power of
Coefficient of x11
1113
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Q
Find the number of terms in different powers of x in:
(a)
(b) (1 - x)201 (1 + x + x2)200
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Q
Find the number of terms in different powers of x in:
(a)
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Solution:
Given,
Clearly we can observe, that every term is giving two new terms in each expansion,
thus n terms will give 2n new terms and a constant term
Thus, total number of terms of different powers of x will be 2n + 1
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Q
Find the number of terms in different powers of x in:
(b) (1 - x)201 (1 + x + x2)200
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Q
Find the number of terms in different powers of x in:
(b) (1 - x)201 (1 + x + x2)200
Solution:
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Do you really understand the meaning of
Expansion or Identity ?
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Q
If (1 - 2x2 + x)6 = a0 + a1x + a2x2 + … + a12x12 , then find:
(a) a0 + a1 + a2 +...+ a12 i.e., sum of coefficients
(b) a0 + a2 + a4 +...+ a12
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Q
If (1 - 2x2 + x)6 = a0 + a1x + a2x2 + … + a12x12 , then find:
(a) a0 + a1 + a2 +...+ a12 i.e., sum of coefficients
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Q
If (1 - 2x2 + x)6 = a0 + a1x + a2x2 + … + a12x12 , then find:
(a) a0 + a1 + a2 +...+ a12 i.e., sum of coefficients
Solution:
Given, (1 - 2x2 + x)6 = a0 + a1x + a2x2 +...+ a12x12
(a) Putting x = 1, we get
a0 + a1 + a2 +...+ a12 = (1 - 2 + 1)6
a0 + a1 + a2 +...+ a12 = 0
….. (1)
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Q
If (1 - 2x2 + x)6 = a0 + a1x + a2x2 + … + a12x12 , then find:
(b) a0 + a2 + a4 +...+ a12
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Q
If (1 - 2x2 + x)6 = a0 + a1x + a2x2 + … + a12x12 , then find:
(b) a0 + a2 + a4 +...+ a12
Solution:
(b) Putting x = –1, we get
Adding (1) and (2), we get
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JEE Main 18th March, 2021
Q
Let (1 + x + 2x2)20 = a0 + a1x + a2x2 + … + a40x40 then
a1 + a3 + a5 + … + a37 is equal to
A
220 (220 - 21)
B
219 (220 - 21)
C
219 (220 + 21)
D
220 (220 + 21)
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JEE Main 18th March, 2021
Q
Let (1 + x + 2x2)20 = a0 + a1x + a2x2 + … + a40x40 then
a1 + a3 + a5 + … + a37 is equal to
A
220 (220 - 21)
B
219 (220 - 21)
C
219 (220 + 21)
D
220 (220 + 21)
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JEE Main 18th March, 2021
Q
Let (1 + x + 2x2)20 = a0 + a1x + a2x2 + … + a40x40 then
a1 + a3 + a5 + … + a37 is equal to
Solution:
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Applications of Binomial Theorem
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Applications of Binomial Theorem
Binomial theorem has some applications in divisibility & remainder problems.
Before we do that, observe that:
(a) (1 + x)n - 1 - nx is divisible by x2
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Applications of Binomial Theorem
Binomial theorem has some applications in divisibility & remainder problems.
Before we do that, observe that:
(a) (1 + x)n - 1 - nx is divisible by x2
(b) For any integer a and b, we say ‘b’ is divisible by ‘a’ if b = ak for some k ∈ Z
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Q
(a) Prove that 62n - 35n - 1 is divisible by 352
(b) Prove that 97 + 79 is divisible by 64
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Q
(a) Prove that 62n - 35n - 1 is divisible by 352
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Solution:
Thus, 62n - 35n - 1 is divisible by 352 for all n ∈ N
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Q
(b) Prove that 97 + 79 is divisible by 64
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Solution:
We have
∴ 97 + 79 is divisible by 64
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Q
If n be a positive integer, prove that 32n + 2 − 8n − 9 is divisible by 64.
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Q
If n be a positive integer, prove that 32n + 2 − 8n − 9 is divisible by 64.
Solution:
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Q
Prove that 132n + 1 + 1 is divisible by 14
Remark
(a) xn - 1 = (x - 1) (xn - 1 + xn - 2 +...+ 1)
(b) If n is odd, then
xn + 1 = (x + 1) (xn - 1 - xn - 2 + xn - 3 -...+ (-1)n - 1)
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Q
Prove that 132n + 1 + 1 is divisible by 14
Solution:
Method I:
132n+1 + 1 = (14 - 1)2n+1 + 1
= 2n+1C0 142n+1 - 2n+1C1(14)2n + 2n+1C2(14)2n+1 + … + 2n+1C2n 14 - 2n+1C2n+1 + 1
= 14[2n+1C0142n - 2n+1C1142n-1 + … + 2n+1C2n] - 1 + 1
= 14 (integer for all n ∈ N)
Thus, 132n + 1 + 1 is divisible by 14 for all n ∈ N
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Q
Prove that 132n + 1 + 1 is divisible by 14
Solution:
Method II:
Since (2n + 1) is odd. Thus,
132n + 1 + 1 = (13 + 1) [132n - 132n - 1 + 132n - 2 +...+ (-1)2n]
= 14 (Integer for all n ∈ N)
Thus 132n + 1 + 1 is divisible by 14 for all n ∈ N
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Q
Remainder when (3)63 is divided by 26, is
A
3
B
24
C
25
D
1
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Q
Remainder when (3)63 is divided by 26, is
A
3
B
24
C
25
D
1
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Q
Remainder when (3)63 is divided by 26, is
Solution:
Here,
⸫ Required remainder is 1
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Q
The sum of the last two digits of (27)27 is
A
2
B
3
C
4
D
5
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Q
The sum of the last two digits of (27)27 is
A
2
B
3
C
4
D
5
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Q
The sum of the last two digits of (27)27 is
Solution:
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Q
If λ is the remainder when 22021 is divided by 17, then
the value of λ must be equal to
A
3
B
7
C
13
D
15
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Q
If λ is the remainder when 22021 is divided by 17, then
the value of λ must be equal to
A
3
B
7
C
13
D
15
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Solution:
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Q
If 5353 - 333 is divided by 10, then the remainder
obtained is _____.
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Q
If 5353 - 333 is divided by 10, then the remainder
obtained is _____.
Ans: 6
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Q
If 5353 - 333 is divided by 10, then the remainder
obtained is _____.
Solution:
xn - yn = (x - y) (xn-1 + xn-2 y + …… + xyn-2 + yn-1)
i.e., xn - yn is divisible by x - y
5353 - 333 = 5353 - 353 + 353 - 333 + 33 - 33
= (5353 - 353) + (353 - 33) - (333 - 33)
Clearly, 1st bracket is divisible by 50 and the last
one is divisible by 30. So, both are divisible by 10.
353 - 33 = 33 ⋅ 350 - 33 = 33 (350 - 1)
= 33((10 - 1)25 - 1) = 27(10 - 1)25 - 27
= 27(10k - 1) - 27 = 270k - 54
= 270 k - 60 + 6 = 10μ + 6.
Hence, the remainder obtained is 6.
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Q
Find last digit of (32)32
Or
Find remainder when (32)32 is divided by 10
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Solution:
Here, (32)32 = (25)32 = 2160
Also,
and so on
Thus after every fourth power, digit in unit place is repeated
Thus digit at unit place for 2160 = 24 × 40 = (24)40
Therefore, unit place digit is 6
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Greatest Term in the expansion of (a + bx)n
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Greatest Term in the expansion of (a + bx)n
To find greatest term in the expansion of (a + bx)n, use the following algorithm
Step 1: Solve
Step 2: If k comes out to be non integer then greatest term is T[k] + 1 ,
where [.] is GIF,
while if k comes out to be an integer then Tk = Tk + 1 both are greatest terms
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Q
(a) Find greatest term in the expansion of (2 + 3x)9 at
(b) Find numerically (i.e. in magnitude) greatest term in
expansion of (3 - 5x)15 at
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Q
(a) Find greatest term in the expansion of (2 + 3x)9 at
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Q
(a) Find greatest term in the expansion of (2 + 3x)9 at
Solution:
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Q
(a) Find greatest term in the expansion of (2 + 3x)9 at
Solution:
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Q
(b) Find numerically (i.e. in magnitude) greatest term in
expansion of (3 - 5x)15 at
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Q
(b) Find numerically (i.e. in magnitude) greatest term in
expansion of (3 - 5x)15 at
Solution:
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Q
If n is an even positive integer greater than 1 and x > 0, then the
condition that the greatest term in the expansion of (1 + x)n may
have the greatest coefficient also is
A
B
C
D
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Q
If n is an even positive integer greater than 1 and x > 0, then the
condition that the greatest term in the expansion of (1 + x)n may
have the greatest coefficient also is
A
B
C
D
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Solution:
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Greatest Term in the expansion of (a + bx)n
NOTE
Among binomial coefficients i.e. nC0, nC1, …, nCn
(a) Middle one i.e.
is greatest if n is even
(b) Two middle coefficients i.e.,
are greatest if n is odd
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R-f Factor Theorem
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R-f Factor Theorem
In this section we will deal with numbers of the form:
Primarily we will be required to comment upon [R] & {R}
[R] will be dented by I
{R} will be denoted by f
Clearly, R = I + f
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R-f Factor Theorem
Approach in these problems will be algorithmic
Step 1: Define G be replacing ‘+’ by ‘–’ sign in the value of R, (0 < G < 1)
Step 2: Either add or subtract G from R so that the value (i.e. RHS is an integer)
Step 3: If G is added, you will always get f + G = 1 while if G is subtracted, you will
always get f = G
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Q
(a) Prove that integral part of
(b) If
, is odd.
where I & f are integral & fractional
parts, then prove that: (I + f)(1 - f) = 1
(c) Prove that the integral part of
and hence if
[ . ] = GIF, prove that RG = 42n + 1.
is even, n ∈ N
and G = R - [R], where
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Q
(a) Prove that integral part of
, is odd.
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Solution:
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Q
(b) If
where I & f are integral & fractional
parts, then prove that: (I + f)(1 - f) = 1
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Solution:
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Solution:
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Q
(c) Prove that the integral part of
and hence if
[ . ] = GIF, prove that RG = 42n + 1.
is even, n ∈ N
and G = R - [R], where
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Solution:
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Solution:
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Binomial Theorem for any Index
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Binomial Theorem for any index
Let ‘n’ be a rational number & ‘x’ be a real number such that |x| < 1,
then:
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Binomial Theorem for any index
Some special cases:
(a)
(b)
(c)
(d)
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Q
Find coefficient of x2 in
, where |3x| < 1.
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Q
Find coefficient of x2 in
Solution:
, where |3x| < 1.
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Q
Find coefficient of x2 in
Solution:
, where |3x| < 1.
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Q
What is the coefficient of x3 in
where |3x| < 1 ?
A
-272
B
-540
C
-870
D
-918
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Q
What is the coefficient of x3 in
where |3x| < 1 ?
A
-272
B
-540
C
-870
D
-918
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Solution:
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Q
JEE Main 25th July, 2021
If b is very small as compared to the value of a, so that the cube
and other higher powers of b/a can be neglected in the identity
then the value of γ is:
A
B
C
D
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Q
JEE Main 25th July, 2021
If b is very small as compared to the value of a, so that the cube
and other higher powers of b/a can be neglected in the identity
then the value of γ is:
A
B
C
D
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Solution:
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Q
If
, then find ‘a’ & ‘b’, given that
Remark
We write
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Q
If
, then find ‘a’ & ‘b’, given that
Solution:
Given,
Comparing both sides, we get
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Multinomial Theorem
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Multinomial Theorem
where r, s & t are non-negative integers
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Multinomial Theorem
where r, s & t are non-negative integers
NOTE
Number of terms in expansion of (x + y + z)n is n + 2C2.
Basically it’s the number of non-negative integer solutions of r + s + t = n
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Q
In the expansion of (x + y +z)17 find
(a) Number of terms
(c) Coefficient of x3y8z2
(b) Coefficient of x4y10z3
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Q
In the expansion of (x + y +z)17 find
(a) Number of terms
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Q
In the expansion of (x + y +z)17 find
(a) Number of terms
Solution:
Number of terms in expansion of (x + y +z)17 is
= 17 + 2C2 = 19C2
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Q
In the expansion of (x + y +z)17 find
(b) Coefficient of x4y10z3
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Q
In the expansion of (x + y +z)17 find
(b) Coefficient of x4y10z3
Solution:
For coefficient of x4y10z3
General term of (x + y +z)17 is given by
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Q
In the expansion of (x + y +z)17 find
(c) Coefficient of x3y8z2
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Q
In the expansion of (x + y +z)17 find
(c) Coefficient of x3y8z2
Solution:
For coefficient of x3y8z2
Here Since 3 + 8 + 2 = 13 ≠ 17
Thus coefficient of x3y8z2 = 0
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JEE Main 10th April 2023
Q
The Coefficient of x7 in (1 - x + 2x3)10 is ________.
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JEE Main 10th April 2023
Q
The Coefficient of x7 in (1 - x + 2x3)10 is ________.
Ans: 960
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Solution:
a
b
c
3
7
0
5
4
1
7
1
2
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Binomial Coefficient Problems
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Binomial Coefficient Problems
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Binomial Coefficient Problems
Recall
(a)(1 + x)n = nC0 + nC1x + nC2x2 +...+ nCnxn
(b)(1 - x)n = nC0 - nC1x + nC2x2 - nC3x3 +...+ (-1)n nCnxn
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Binomial Coefficient Problems
(a)(1 + x)n = nC0 + nC1x + nC2x2 +...+ nCnxn
Putting x = 1, we get
2n = nC0 + nC1 + nC2 +...+ nCn
i.e. sum of binomial coefficients is 2n
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Binomial Coefficient Problems
(a)(1 + x)n = nC0 + nC1x + nC2x2 +...+ nCnxn
Putting x = 1, we get
2n = nC0 + nC1 + nC2 +...+ nCn
i.e. sum of binomial coefficients is 2n
(a)(1 - x)n = nC0 - nC1x + nC2x2 - nC3x3 +...+(-1)n nCnxn
Putting x = 1, we get
i .e. 0 + nC0 - nC1 + nC2 - nC3 +...+ (-1)n nCn
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Binomial Coefficient Problems
We majorly categorize the problems in two categories
Category 1:
(a) n & r both change
(b) Only r change
(c) Only n change
Category 2:
Result based Questions
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Q
Find the value of
(a) 50C25 + 49C24 +...+ 26C1 + 25C0
(b) 15C8 - 14C8 - 13C7 - 12C6 - 11C5
(c) 50C10 + 51C10 + 52C10 +...+ 100C10
(d) 23C0 + 23C2 + 23C4 +...+ 23C22
(e) 20C0 - 20C1 + 20C2 - 20C3 + … - … + 20C10
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Q
Find the value of : (a) 50C25 + 49C24 +...+ 26C1 + 25C0
Recall
nC
r+
nC
r+1 =
n + 1C
r + 1 (or)
n + 1C
r+1 -
nC
r+1 =
nC
r
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Q
Find the value of : (a) 50C25 + 49C24 +...+ 26C1 + 25C0
Solution:
Given 50C25 + 49C24+...+ 26C2 + 26C1 + 25C0
⇒ 25C0 + 26C1 +...+ 49C24 + 50C25
⇒ (26C0 + 26C1) +...+ 49C24 + 50C25
⇒ (27C1 + 27C2) +...+ 49C24 + 50C25
⇒ 51C25
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Q
Find the value of: (b) 15C8 - 14C8 - 13C7 - 12C6 - 11C5
Recall
nC
r+
nC
r+1 =
n + 1C
r + 1 (or)
n + 1C
r+1 -
nC
r+1 =
nC
r
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Q
Find the value of: (b) 15C8 - 14C8 - 13C7 - 12C6 - 11C5
Solution:
Given (15C8 - 14C8) - 13C7 - 12C6 - 11C5
⇒(14C7 - 13C7) - 12C6 - 11C5
= (13C6 - 12C6) - 11C5
= 12C5 - 11C5
= 11C4
[ n + 1Cr + 1 - nCr + 1 = nCr]
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Q
Find the value of: (c) 50C10 + 51C10 + 52C10 +...+ 100C10
Recall
nC
r+
nC
r+1 =
n + 1C
r + 1 (or)
n + 1C
r+1 -
nC
r+1 =
nC
r
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Solution:
When only ‘n’ changes we normally proceed as follows:
It is the coefficient of x10 in (1 + x)50 + (1 + x)51 + (1 + x)52 +...+ (1 + x)100
coefficient of x10 in
coefficient of x10 in
coefficient of x10 is 101C11 - 50C11
50C
10 +
51C
10 +
52C
10 +...+
100C
10 =
101C
11 -
50C
11
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Alternate Solution
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Q
Find the value of: (d) 23C0 + 23C2 + 23C4 +...+ 23C22
Recall
(a) nC0 + nC1 + nC2 +...+ nCn = 2n
(b) nC0 + nC2 + nC4 +... = nC1 + nC3 + nC5 +... = 2n-1
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Q
Find the value of: (d) 23C0 + 23C2 + 23C4 +...+ 23C22
Solution:
Here we have all the even binomial
coefficients with n = 23
23C
0+
23C
2+
23C
4 +...
23C
22 = 2
n-1
= 223 - 1
= 222
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Q
Find the value of: (e) 20C0 - 20C1 + 20C2 - 20C3 + … - … + 20C10
Recall
(a) nC0 + nC1 + nC2 +...+ nCn = 2n
(b) nC0 + nC2 + nC4 +... = nC1 + nC3 + nC5 +... = 2n-1
Slide 181
2
iska sol nhin hai kya?
1
Sir, Nhi tha pehle, abhi add kiya hai
Arvind Kalia, 25-07-2023
anurag tiwari, 25-07-2023
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Q
Find the value of: (e) 20C0 - 20C1 + 20C2 - 20C3 + … - … + 20C10
Solution:
Given 20C0 - 20C1 + 20C2 - 20C3 + . . . - . . . + 20C10
Since 20C0 - 20C1 + 20C2 - . . . - 20C9 + 20C10 - 20C11 + . . . + 20C20 = 0
⇒ 2[20C0 - 20C1 + . . . - 20C9 + 20C10 ] = 20C10
⇒ [20C0 - 20C1 + . . . - 20C9 + 20C10 ] = ½ x 20C10
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Q
The value of
is equal to
A
B
C
D
128
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Q
The value of
is equal to
A
B
C
D
128
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Q
The value of
Solution:
is equal to
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Q
JEE Advanced 2016, P1
Let m be the smallest positive integer such that the coefficient of
x2 in the expansion of (1 + x)2 + (1 + x)3+ … + (1 + x)49 + (1 + mx)50 is
(3n + 1) 51C3 for some positive integer n. Then the value of n is
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Q
JEE Advanced 2016, P1
Let m be the smallest positive integer such that the coefficient of
x2 in the expansion of (1 + x)2 + (1 + x)3+ … + (1 + x)49 + (1 + mx)50 is
(3n + 1) 51C3 for some positive integer n. Then the value of n is
Ans: 5
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Q
JEE Advanced 2016, P1
Let m be the smallest positive integer such that the coefficient of
x2 in the expansion of (1 + x)2 + (1 + x)3+ … + (1 + x)49 + (1 + mx)50 is
(3n + 1) 51C3 for some positive integer n. Then the value of n is
Solution:
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Binomial Coefficient Problems
Result
(1)
(2)
(3)
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Binomial Coefficient Problems
Result
(4)
(5)
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JEE Main 25th July, 2021
Q
Let n ∈ N and [x] denote the greatest integer less than or
equal to x. If the sum of (n + 1) terms nC0 , 3 .nC1 , 5 .nC2 , 7 .nC3 …
is equal to 2100 . 101 then
is equal to_____
Hint:
(1)
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JEE Main 25th July, 2021
Q
Let n ∈ N and [x] denote the greatest integer less than or
equal to x. If the sum of (n + 1) terms nC0 , 3 .nC1 , 5 .nC2 , 7 .nC3 …
is equal to 2100 . 101 then
Ans: 98
is equal to_____
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Solution:
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JEE Advanced 2018, P2
Q
Let X = (10C1)2 + 2(10C2)2 + 3(10C3)2 + … + 10(10C10)2, where
10C , r ∈ {1, 2, …, 10} denote binomial coefficients. Then,
r
the value of
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JEE Advanced 2018, P2
Q
Let X = (10C1)2 + 2(10C2)2 + 3(10C3)2 + … + 10(10C10)2, where
10C , r ∈ {1, 2, …, 10} denote binomial coefficients. Then,
r
the value of
Ans: 646
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Solution:
(1 + x)10 = 10C0 + 10C1x + 10C2x2 + … + 10C10x10
Differentiate both sides w.r.t. x,
10(1 + x)9 = 10C1 + 2 ⋅ 10C2x + 310C3x2 + … + 1010C10x9
…(1)
(1 + x)10 = 10C0x10 + 10C1x9 + … + 10C10
…(2)
Coefficient of x9 in 10(1 + x)9 (1 + x)10 is same as
(10C1)2 + 2(10C2)2 + 3(10C3)2 + … + 10(10C10)2
= Coefficient of x9 in 10(1 + x)19 = 10 × 19C9 = X
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Q
Find the sum of
10C
0⋅
10C
3+
10C ⋅ 10C + 10C ⋅ 10C + … + 10C ⋅ 10C
1
4
2
5
7
10
Hint:
(4)
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Q
Find the sum of
10C
0⋅
10C
3+
Solution:
10C ⋅ 10C + 10C ⋅ 10C + … + 10C ⋅ 10C
1
4
2
5
7
10
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JEE Main 28th June, 2022
Q
Where ⍺ ∈ R, then the value of 16⍺ is equal to
A
1411
B
1320
C
1615
D
1855
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JEE Main 28th June, 2022
Q
Where ⍺ ∈ R, then the value of 16⍺ is equal to
A
1411
B
1320
C
1615
D
1855
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Solution:
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IIT JEE 2005
Q
The value of
A
B
C
D
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IIT JEE 2005
Q
The value of
A
B
C
D
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Solution: