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Test: Binomial Theorem- 1 - JEE MCQ


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25 Questions MCQ Test Physics for JEE Main & Advanced - Test: Binomial Theorem- 1

Test: Binomial Theorem- 1 for JEE 2025 is part of Physics for JEE Main & Advanced preparation. The Test: Binomial Theorem- 1 questions and answers have been prepared according to the JEE exam syllabus.The Test: Binomial Theorem- 1 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Binomial Theorem- 1 below.
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Test: Binomial Theorem- 1 - Question 1
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If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are

Detailed Solution for Test: Binomial Theorem- 1 - Question 1

(x+a)n = nC0 xn + nC1 x(n-1) a1 + nC2 x(n-2) a2 + ..........+ nC(n-1) xa(n-1) + nCn  an
Now, nC0 = nCn, nC1 = nCn-1,    nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

Test: Binomial Theorem- 1 - Question 2
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If the coefficients of x−7 and x−8 in the expansion of are equal then n =

Detailed Solution for Test: Binomial Theorem- 1 - Question 2

coefficient of x-7 in [2+1/3x]n is nC7(2)n-7 (1/3)7
coefficient of x-8 in [2+1/3x]n is nC8(2)n-8 (1/3)8
⟹ nC7 × (2n-7) × 1/37 = nC8 × (2n-8) × 1/38
⟹ nC8/nC7 = 6
⟹ (n-7)/8 = 6
⟹ n = 55

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Test: Binomial Theorem- 1 - Question 3
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If 2nd, 3rd and 4th terms in the expansion of (x+a)n are 240, 720 and 1080 respectively, then the value of n is

Detailed Solution for Test: Binomial Theorem- 1 - Question 3

General term Tr+1 of (x+y)n is given by 
Tr+1 = nCr xn-r yr
T2 = nC2 xn-2 y = 240
T3 = nC3 xn-3 y2 = 720
T4 = nC4 xn-4 y3 = 1080
T3/T2 = [(n-1)/2] * [y/x] = 3......(1)
T4/T2 = {[(n-1)(n-2)]/(3*2)} * x2/y2 = 9/2
T4/T3 = [(n-2)/3] * [y/x] = 3/2...(2)
Dividing 1 by 2
[(n-1)/2] * [3/(n-2)] = 2
⇒ 3n−3 = 4n−8
⇒ 5 = n

Test: Binomial Theorem- 1 - Question 4
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The coefficient of xn in expansion of (1+x)(1−x)n is
Detailed Solution for Test: Binomial Theorem- 1 - Question 4

We can expand (1+x)(1−x)n as 
= (1+x)(nC0​− nC1​.x + nC2​.x2 + ........ +(−1)n nCn​xn)
Coefficient of xn is 
(−1)n nCn ​+ (−1)n−1 nCn−1
​=(−1)n(1−n)

Test: Binomial Theorem- 1 - Question 5
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The coefficient of x70 in
x2(1 + x)98 + x3(1 + x)97 + x4(1 + x)96 + ... + x54(1 + x)46 is
99Cp - 46Cq. Then a possible value of p + q is:
Detailed Solution for Test: Binomial Theorem- 1 - Question 5

x²(1 + x)⁹⁸ + x³(1 + x)⁹⁷ + ... + x⁵⁴(1 + x)⁴⁶
It is a G.P. with first term = x²(1 + x)⁹⁸
and common ratio = x / (1 + x)
sum of these terms = x²(1 + x)⁹⁸ * [( (x / (1 + x))⁵³ - 1 ) / (x / (1 + x) - 1))]
= x²(1 + x)⁹⁸ * ((1 + x) - x⁵³(1 + x)⁻⁵²)

= ⁹⁹C₆₈ - ⁴⁶C₁₅
⇒ p = 68, q = 15
⇒ p + q = 83

Test: Binomial Theorem- 1 - Question 6
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The coefficient of y in the expansion of (y² + c/y)5 is 
Detailed Solution for Test: Binomial Theorem- 1 - Question 6

Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

Test: Binomial Theorem- 1 - Question 7
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If the term independent of x in the expansion of (√(ax2) + 1/2x3)10 is 105 , then a2 is equal to:
Detailed Solution for Test: Binomial Theorem- 1 - Question 7

(√a x2 + (1 / 2x3))10
Tr+1 = ¹⁰Cr (√a x2)10-r (1 / 2x3)r
Independent of x ⇒ 20 - 2r - 3r = 0
r = 4
Independent of x is ¹⁰C₄ (√a)6 (1 / 2)4 = 105
(210 / (2 × 8)) a3 = 105
⇒ a = 2
a2 = 4

Test: Binomial Theorem- 1 - Question 8
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If the constant term in the expansion of( (5√3) / x + (2x) / 3√5 )12 , x ≠ 0, is α × 2⁸ × ⁵√3, then 25α is equal to:
Detailed Solution for Test: Binomial Theorem- 1 - Question 8


For constant term - 12 + r + r = 0
⇒ r = 6
∴ Constant term = 
= ¹²C₆ × (2⁶ / 25) × 3 × 31/5
= (231 / 25) × 2⁸ × 31/5 × 3
= (693 / 25) × 2⁸ × ⁵√3
∴ α = 693 / 25
25α = 693

Test: Binomial Theorem- 1 - Question 9
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The coefficient of y in the expansion of 
Detailed Solution for Test: Binomial Theorem- 1 - Question 9

5C0 (c/y)0 (y2)5-0 + 5C1 (c/y)1 (y2)5-1 + 5C2 (c/y)2 (y2)5-2 +....... + 5C5 (c/y)5 (y2)5-5
∑(r = 0 to 5) 5Cr (c/y)r (y2)5-r
We need coefficient of y ⇒2(5−r)−r=1
⇒ 10 − 3r = 1
⇒ r = 3
So, cofficient of y = 5C3.C3
= 10C3

Test: Binomial Theorem- 1 - Question 10
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If the coefficients of x⁴, x⁵ and x⁶ in the expansion of (1 + x)ⁿ are in the arithmetic progression, then the maximum value of n is:
Detailed Solution for Test: Binomial Theorem- 1 - Question 10

(1 + x)ⁿ = ⁿC₀ + ⁿC₁x¹ + ⁿC₂x² + ... + ⁿCₙxⁿ
ⁿC₄, ⁿC₅ & ⁿC₆ are in A.P.
ⁿC₅ - ⁿC₄ = ⁿC₆ - ⁿC₅
⇒ n! / (5!(n - 5)!) - n! / (4!(n - 4)!) = n! / (6!(n - 6)!) - n! / (5!(n - 5)!)
⇒ 30(n - 9)(n - 6) = 5(n - 4)(n - 11)
⇒ 30n² - 450n + 1620 = 5n²
⇒ 1 / (n - 5) [ (n - 4 - 5) / (5(n - 4)) ] = 1 / 5 [ (n - 5 - 6) / (6(n - 5)) ]
⇒ (n - 9) / (5(n - 4)) = 1 / 5 [ (n - 11) / 6 ]
⇒ n² - 21n + 98 = 0
nₘₐₓ = 14

Test: Binomial Theorem- 1 - Question 11
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The sum of all rational terms in the expansion of (21/5 + 51/3)15 is equal to:
Detailed Solution for Test: Binomial Theorem- 1 - Question 11


For rational terms,
r/3 and r/5 must be integer
3 and 5 divide r ⇒ 15 divides r ⇒ r = 0 and r = 15
¹⁵C₀ 5⁰ 2³ + ¹⁵C₁₅ 5⁵ 2⁰
= 8 + 3125
= 3133

Test: Binomial Theorem- 1 - Question 12
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Suppose 2 - p, p, 2 - α, α are the coefficients of four consecutive terms in the expansion of (1 + x)ⁿ.Then the value of p² - α² + 6α + 2p equals
Detailed Solution for Test: Binomial Theorem- 1 - Question 12

Given that 2 - p, p, 2 - α, α are four consecutive terms in the binomial expansion of (1 + x)ⁿ, they must follow the property of binomial coefficients:
Next term / Previous term = (n - r) / (r + 1)
Using this property:
(p / (2 - p)) = ((n - (r - 1)) / r)
((2 - α) / p) = ((n - r) / (r + 1))
(α / (2 - α)) = ((n - (r + 1)) / (r + 2))
After solving these equations, it turns out that the given expression:
p² - α² + 6α + 2p
simplifies to 8.
Thus, the correct option is: A. 8.

Test: Binomial Theorem- 1 - Question 13
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n-1Cr = (k² - 8) nCr+1 if and only if:
Detailed Solution for Test: Binomial Theorem- 1 - Question 13

n-1Cr = (k² - 8) nCr+1

(n-1Cr) / (nCr+1) = k² - 8
(r + 1) / n = k² - 8
⇒ k² - 8 > 0
(k - 2√2)(k + 2√2) > 0
k ∈ (-∞, -2√2) ∪ (2√2, ∞) .... (I)
∴ n ≥ r + 1, (r + 1) / n ≤ 1
⇒ k² - 8 ≤ 1
k² - 9 ≤ 0
-3 ≤ k ≤ 3 .... (II)
From equation (I) and (II) we get
k ∈ [-3, -2√2) ∪ (2√2, 3]

Test: Binomial Theorem- 1 - Question 14
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Coefficient of x5 in the expansion of (1+x2)5(1+x)4 is  
Detailed Solution for Test: Binomial Theorem- 1 - Question 14

Co-efficient of x5 in (1+x2)5 (1+x)4
= {5C0 (1)0 + 5C1 (1)x2 + 5C2 (1)(x2)2 + 5C3 (1)(x2)3 + 5C4 (1)(x2)4 + 5C5 (1)(x2)5} * {4C0 (1) + 4C1 (x) + 4C2 (x)2 + 4C3 (x)3 + 4C4 (x)4}
= {5C2(x)4 * 4C1(x) + 5C1(x)4 + 5C1(x)2 * 4C3(x)3} 
∴ coefficient of x5
=> (5×4)/(2×1) × 4)x5 + (5×4)x5
⇒(10×4+20)x5
⇒ 60x5

Test: Binomial Theorem- 1 - Question 15
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If A denotes the sum of all the coefficients in the expansion of (1 - 3x + 10x²)ⁿ and B denotes the sum of all the coefficients in the expansion of (1 + x²)ⁿ, then:
Detailed Solution for Test: Binomial Theorem- 1 - Question 15

Sum of coefficients in the expansion of (1 - 3x + 10x²)ⁿ = A
then A = (1 - 3 + 10)ⁿ = 8ⁿ (put x = 1)
and sum of coefficients in the expansion of
(1 + x²)ⁿ = B
then B = (1 + 1)ⁿ = 2ⁿ
A = B³

*Answer can only contain numeric values
Test: Binomial Theorem- 1 - Question 16
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If the second, third, and fourth terms in the expansion of (x + y)n are 135, 30, and 10/3, respectively, then 6 (n3 + x2 + y) is equal to ________.

Detailed Solution for Test: Binomial Theorem- 1 - Question 16

T2 = nC1 y1 x(n-1) = 135
T3 = nC2 y2 x(n-2) = 30
T4 = nC3 y3 x(n-3) = 10/3
⇒ 135/30 = (x/y) * n * 2 / n(n-1) = (2 / n-1) * (x/y) ... (i)
30 / (10/3) = n(n-1) / 2 / n(n-1)(n-2) * 3! * (x/y)
9 = (3 / n-2) * (x/y)
3(n - 2) = 135 / 60 (n - 1) ⇒ n = 5
⇒ x = 9y ... (i)
y * x4 = 27 ⇒ x / 9 * x4 = 33
⇒ x5 = 35 ⇒ x = 3y = 1/3
⇒ 6 (53 + 32 + 1/3) = 6 (125 + 9 + 1/3)
= 6(134) + 2 = 806

Test: Binomial Theorem- 1 - Question 17
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The coefficient of x⁵ in the expansion of (2x³ - (1 / 3x²))⁵ is:
Detailed Solution for Test: Binomial Theorem- 1 - Question 17

Given, (2x³ - (1 / 3x²))⁵
General term,

∴ 15 - 5r = 5
∴ r = 2
T₃ = 10 (8 / 9) x⁵
So, coefficient is 80 / 9.

*Answer can only contain numeric values
Test: Binomial Theorem- 1 - Question 18
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If the constant term in the expansion of (1 + 2x - 3x3)(3/2 x2 - 1/3 x)9 is p, then 108p is equal to_______.

Detailed Solution for Test: Binomial Theorem- 1 - Question 18

General term of (3/2 x2 - 1/3 x)9
Tr+1 = 9Cr(3/2 x2)(9-r)(-1/3x)r = 9Cr(-1)r39-2r2r-9x18-3
Constant term in expansion of (1 + 2x - 3x3)
(3/2 x2 - 1/3 x)9
= T7 - T8 = 9C6 3(-3) 2(-3) + 39C7 3(-5) 2(-2)
= 3 × 4 × 7 / 33 * 23 + 3 × 9 × 4 / 35 × 22
p = 54 / 108
108p = 54

Test: Binomial Theorem- 1 - Question 19
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Fractional part of the number (4²⁰²² / 15) is equal to
Detailed Solution for Test: Binomial Theorem- 1 - Question 19

{ (42022) / 15 } = { (24044) / 15 }
= { (1 + 15)1011 / 15 }
= 1 / 15

*Answer can only contain numeric values
Test: Binomial Theorem- 1 - Question 20
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Let the coefficient of x^r in the expansion of
(x + 3)n-1 + (x + 3)n-2(x + 2) + (x + 3)n-3(x + 2)2 + .......... + (x + 2)n-1 be αr. If
∑ (from r = 0 to n) αr = βn - γn, β, γ ∈ ℕ, then the value of β2 + γ2 equals ________."

Detailed Solution for Test: Binomial Theorem- 1 - Question 20

(x + 3)(n-1) + (x + 3)(n-2)(x + 2) + (x + 3)(n-3)(x + 2)2 + ... + (x + 2)(n-1)
∑ αr = 4(n-1) + 4(n-2) × 3 + 4(n-3) × 32 + ... = 4(n-1) [1 + 3/4 + (3/4)2 + ... + (3/4)(n-1)]
= 4(n-1) × (1 - (3/4)n) / (1 - 3/4)
= 4(n-1) × 4 × (1 - (3/4)n)
= 4(n-1) × 4 × (βn - γn)
β = 4, γ = 3
β2 + γ2 = 16 + 9 = 25

Test: Binomial Theorem- 1 - Question 21
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5th term from the end in the expansion of 
Detailed Solution for Test: Binomial Theorem- 1 - Question 21

5th term in the expansion of (x2/2 − 2/x2)12 is
Tr+ 1= nCr xr* y(n−r)
T5 = 12C4[(x2/2)4 (-2/x2)8]
= (12! * x8 * 28)/ (4! * 8! *24 * x16)
= 7920x−4

Test: Binomial Theorem- 1 - Question 22
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The sum of the coefficient of x2/3 and x-2/5 in the binomial expansion of (x2/3 + (1/2)x-2/5)9 is
Detailed Solution for Test: Binomial Theorem- 1 - Question 22


For coefficient of x2/3
⇒ 6 - (16r/15) = 2/3
⇒ 90 - 16r = 10
⇒ r = 5
For coefficient of x-2/5
⇒ 6 - (16r/15) = -2/5
⇒ 90 - 16r = -6
⇒ r = 6
Sum of coefficient of x2/3 & x-2/5
= ⁹C₅ . (1/25) + ⁹C₆ . (1/26)
= (9! / 5!4!) x (1/25) + (9! / 6!3!) x (1/26) = 21/4

Test: Binomial Theorem- 1 - Question 23
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The sum of the coefficients of the first 50 terms in the binomial expansion of (1 - x)¹⁰⁰, is equal to
Detailed Solution for Test: Binomial Theorem- 1 - Question 23

(¹⁰⁰C₀ - ¹⁰⁰C₁ + ¹⁰⁰C₂ - ... - ¹⁰⁰C₄₉) + ¹⁰⁰C₅₀
+ (-¹⁰⁰C₅₁ + ¹⁰⁰C₅₂ - ... + ¹⁰⁰C₁₀₀) = 0
λ₁ + ¹⁰⁰C₅₀ + λ₂ = 0
λ₁ = - (1/2) ¹⁰⁰C₅₀ (∵ λ₁ = λ₂)
= -⁹⁹C₄₉

*Answer can only contain numeric values
Test: Binomial Theorem- 1 - Question 24
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The remainder when 4282024 is divided by 21 is ________.

Detailed Solution for Test: Binomial Theorem- 1 - Question 24

428 = 21 × 20 + 8
⇒ 4282024 ≡ (20 × 21 + 8)2024 ≡ 82024 (mod 21)
82 = 21 × 3 + 1
82024 = (21 × 3 + 1)1012
⇒ 82024 ≡ (21 × 3 + 1)1012 (mod 21)
≡ 12012 (mod 21)
4282024 ≡ 1 (mod 21)

*Answer can only contain numeric values
Test: Binomial Theorem- 1 - Question 25
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Number of integral terms in the expansion of (71/2 + 111/6)824 is equal to ________.

Detailed Solution for Test: Binomial Theorem- 1 - Question 25

General term in expansion of (71/2 + 111/6)824 is T(r+1) = 824Cr * 7((824-r)/2) * 11(r/6)
For integral term, r must be a multiple of 6.
Hence r = 0, 6, 12, ..., 822

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