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Binomial Theorem appeared 37 times across 3 years — 4.3% of Mathematics. This question is from Binomial Coefficients Series.

Year 2026 2025 2024 Total
Questions 9 17 11 37

If Σr=1⁹( r + 32r)·⁹Cᵣ = α ((3)/(2))⁹ - β [cite: 596], where α, β in N [cite: 596], then (α + β)² is equal to[cite: 608]:

Solution & Explanation

Related Formula
  • Σr=1ⁿ r · ⁿCᵣ x^r = nx(1+x)ⁿ⁻¹
  • Binomial Theorem expansion: Σr=0ⁿ ⁿCᵣ x^r = (1+x)ⁿ
Core Logic

Split the given series summation into two independent parts [cite: 1316]: Sum = Σr=1⁹ (r)/(2^r) · ⁹Cᵣ + 3Σr=1⁹ (1)/(2^r) · ⁹Cᵣ [cite: 1316]

Simplify the first sub-sum using the index relation r · ⁹Cᵣ = 9 · ⁸Cᵣ₋₁ [cite: 1316]: Σr=1⁹ (9)/(2^r) · ⁸Cᵣ₋₁ = (9)/(2)Σr=1⁹ ⁸Cᵣ₋₁((1)/(2))r-1 = (9)/(2)(1 + (1)/(2))⁸ = (9)/(2)((3)/(2))⁸ [cite: 1316]

Simplify the second sub-sum by including missing index r=0 [cite: 1316]: 3[Σr=0⁹ ⁹Cᵣ((1)/(2))^r - 1] = 3[(1 + (1)/(2))⁹ - 1] = 3((3)/(2))⁹ - 3 [cite: 1316]

Step 1: Combining the components

Combine both evaluations to fit into requested representation shape [cite: 1316]: Total Sum = (9)/(2)((3)/(2))⁸ + 3((3)/(2))⁹ - 3 [cite: 1316] Convert the fractional leading term [cite: 1316]: (9)/(2)((3)/(2))⁸ = 3 · (3)/(2)((3)/(2))⁸ = 3((3)/(2))⁹ [cite: 1316] Total Sum = 3((3)/(2))⁹ + 3((3)/(2))⁹ - 3 = 6((3)/(2))⁹ - 3 [cite: 1316]

Matching coefficients gives [cite: 1317]: α = 6, β = 3 [cite: 1317]

Evaluate the required squared value [cite: 1317]: (α + β)² = (6 + 3)² = 81 [cite: 1317]

Pattern Recognition

Splitting variable factors into standard combinatoric property fractions simplifies coefficient conversions with geometric series denominators.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 6

Q58 jee_main_2025_29_jan_morning Number of Integral Terms
The least value of n for which the number of integral terms in the Binomial expansion of (3√(7) + 12√(11))ⁿ is 183, is:
  • A. 2184
  • B. 2148
  • C. 2172
  • D. 2196

Solution

Related Formula
Tᵣ₊₁ = nr an-r b^r
Core Logic

The general term in the expansion is:

Tᵣ₊₁ = nr (7)(n-r)/(3) (11)(r)/(12)

For the term to be integral, both powers (n-r)/(3) and (r)/(12) must be integers. This dictates that r must be a multiple of 12 (r = 12k, where k is a non-negative integer).

Step 1: Setup total count equation

The values r can take are 0, 12, 24, , 12k. The total number of terms is given as 183. Since counting starts from k=0, the maximum value of k is:

k = 183 - 1 = 182
Step 2: Calculate minimum n

The maximum index value r required to achieve this count is:

r = 12 × 182 = 2184

Hence, the least value of n must be 2184 to encompass all 183 integral terms.

Pattern Recognition

Number of terms formula when tracking steps of size L: Total Terms = (n)/(L) + 1. Isolate n directly to compute bounds rapidly.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q23 jee_main_2024_01_february_morning General Term and Coefficients
If the Coefficient of x³⁰ in the expansion of (1+(1)/(x))⁶(1+x²)⁷(1-x³)⁸, x≠0 is α, then |α| equals
Numerical Answer. Answer: 678 to 678

Solution

Related Formula

General term in a binomial expansion (1+t)ⁿ is given by:

Tᵣ₊₁ = nr t^r
Core Logic

Let's simplify the algebraic structure of the product expression first:

(1+(1)/(x))⁶(1+x²)⁷(1-x³)⁸ = ((x+1)⁶ (1+x²)⁷ (1-x³)⁸)/(x⁶)

Finding the coefficient of x³⁰ in this full product is equivalent to finding the coefficient of x³⁶ in the numerator expansion:

Target = Coefficient of x³⁶ in (1+x)⁶ (1+x²)⁷ (1-x³)⁸
Step 1: Setting up General Term Constraints

The product of the three general terms is:

6r₁ xr₁ · 7r₂ (x²)r₂ · 8r₃ (-x³)r₃ = 6r₁ 7r₂ 8r₃ (-1)r₃ xr₁ + 2r₂ + 3r₃

We require the total exponent to equal 36:

r₁ + 2r₂ + 3r₃ = 36

with boundaries 0 ≤ r₁ ≤ 6, 0 ≤ r₂ ≤ 7, 0 ≤ r₃ ≤ 8.

Step 2: Case Analysis by r3

Let's evaluate non-vanishing integer combinations case-by-case:

  • Case I: r₃ = 8
r₁ + 2r₂ = 36 - 24 = 12
  • r₂ = 6, r₁ = 0 60 76 88(-1)⁸ = 1 × 7 × 1 = 7
  • r₂ = 5, r₁ = 2 62 75 88(-1)⁸ = 15 × 21 × 1 = 315
  • r₂ = 4, r₁ = 4 64 74 88(-1)⁸ = 15 × 35 × 1 = 525
  • r₂ = 3, r₁ = 6 66 73 88(-1)⁸ = 1 × 35 × 1 = 35
  • Case II: r₃ = 7
r₁ + 2r₂ = 36 - 21 = 15
  • r₂ = 7, r₁ = 1 61 77 87(-1)⁷ = 6 × 1 × 8 × (-1) = -48
  • r₂ = 6, r₁ = 3 63 76 87(-1)⁷ = 20 × 7 × 8 × (-1) = -1120
  • r₂ = 5, r₁ = 5 65 75 87(-1)⁷ = 6 × 21 × 8 × (-1) = -1008
  • Case III: r₃ = 6
r₁ + 2r₂ = 36 - 18 = 18
  • r₂ = 7, r₁ = 4 64 77 86(-1)⁶ = 15 × 1 × 28 = 420
  • r₂ = 6, r₁ = 6 66 76 86(-1)⁶ = 1 × 7 × 28 = 196
Step 3: Summation for Alpha

Summing all calculated values:

α = (7 + 315 + 525 + 35) + (-48 - 1120 - 1008) + (420 + 196) α = 882 - 2176 + 616 = -678

Thus, the absolute value is:

|α| = 678

Pattern Recognition

Sees: Multi-product polynomial coefficient extraction problem. Trap: Remember that the negative sign inside (1-x³)⁸ alters the polarity of terms based on whether r₃ is odd or even. Always track the (-1)r₃ factor carefully.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q28 jee_main_2024_29_january_evening Remainder and Divisibility Problems
Remainder when 64^32³² is divided by 9 is equal to
Numerical Answer. Answer: 1 to 1

Solution

Related Formula
(9k + 1)ⁿ ≡ 1 9
Core Logic

Let the giant power tower expression be variable parameter t = 32³². Re-expressing baseline number base 64: 64 = 8² So, the total term evaluates to:

64^t = (8²)^t = 82t
Step 1: Evaluating Modulo Properties

We can re-express baseline 8 as (9 - 1):

82t = (9 - 1)2t

Expanding this expression using the Binomial Theorem:

(9 - 1)2t = 2t092t - + 2t2t(-1)2t

Since the exponent 2t is an even number, (-1)2t = 1. Therefore:

(9 - 1)2t = 9k + 1 ≡ 1 9

Thus, the remainder obtained when divided by 9 is exactly 1.

Pattern Recognition

When evaluating exponents modulo M, reduce the base first. Since 64 ≡ 1 9, any integer power of 64 trivially leaves a remainder of 1.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q1 jee_main_2024_27_jan_morning Properties of Binomial Coefficients
Given ⁿ⁻¹Cᵣ=(k²-8)ⁿCᵣ₊₁, this holds true if and only if:
  • A. 2√(2) lt k ≤ 3
  • B. 2√(3) lt k ≤ 3√(2)
  • C. 2√(3) lt k lt 3√(3)
  • D. 2√(2) lt k lt 2√(3)

Solution

Related Formula
ⁿ⁻¹CᵣⁿCᵣ₊₁ = (r+1)/(n)
Core Logic

From the given equation:

(k² - 8) = ⁿ⁻¹CᵣⁿCᵣ₊₁

Applying the combination property, we get:

k² - 8 = (r+1)/(n)

Since n, r ≥ 0 and n ≥ r+1 for the combination to be valid, the ratio (r+1)/(n) must satisfy:

0 < (r+1)/(n) ≤ 1

Substituting this bound into our expression:

0 < k² - 8 ≤ 1
Step 1: Solving the Inequalities

First inequality:

k² - 8 > 0 ⇒ k² > 8 ⇒ k in (-∞, -2√(2)) (2√(2), ∞)

Second inequality:

k² - 8 ≤ 1 ⇒ k² - 9 ≤ 0 ⇒ k² ≤ 9 ⇒ -3 ≤ k ≤ 3

Taking the intersection of both intervals:

k in [-3, -2√(2)) (2√(2), 3]
Step 2: Final Conclusion

Looking at the options, we consider the positive domain interval:

2√(2) < k ≤ 3
Pattern Recognition

Combinatorics identities often reduce to bounds on variables. Memorize the ratio ⁿ⁻¹CᵣⁿCᵣ₊₁ = (r+1)/(n) and use the strict combinatorial limit 0 < (r+1)/(n) ≤ 1 to form algebraic inequalities.

Chapter Mix

Class 11 Maths: Binomial Theorem Class 11 Maths: Linear Inequalities

Q5 jee_main_2024_27_jan_morning Sum of Binomial Coefficients
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 :
  • A. A=B³
  • B. 3A=B
  • C. B=A³
  • D. A=3B

Solution

Related Formula
Sum of all coefficients in f(x)ⁿ = f(1)ⁿ
Core Logic

To find the sum of all coefficients in a polynomial expansion, substitute the variable x = 1.

For A (Sum of coefficients of (1-3x+10x²)ⁿ):

A = (1 - 3(1) + 10(1)²)ⁿ A = (1 - 3 + 10)ⁿ = (8)ⁿ

For B (Sum of coefficients of (1+x²)ⁿ):

B = (1 + (1)²)ⁿ B = (1 + 1)ⁿ = (2)ⁿ
Step 1: Establishing the Relation

We have A = 8ⁿ and B = 2ⁿ. Observe that 8 = 2³.

A = (2³)ⁿ = (2ⁿ)³

Substituting B = 2ⁿ into the expression: A = B³

Pattern Recognition

Summing coefficients of any multi-nomial expansion is trivially simple: just plug in x=1 (or x=y=z=1 for multivariable expressions). It collapses the variables, leaving only the arithmetic sum of the coefficients.

Chapter Mix

Class 11 Maths: Binomial Theorem

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