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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 8

Q2 jee_main_2024_31_jan_morning Sum of Coefficients and Limits
Let a be the sum of all coefficients in the expansion of (1 - 2x + 2x²)²⁰²³ (3 - 4x² + 2x³)²⁰²⁴ and b = x → 0 ( ∫₀x (1 + t)t²⁰²⁴ + 1 dtx² ). If the equations cx² + dx + e = 0 and 2bx² + ax + 4 = 0 have a common root, where c, d, e in R, then d : c : e equals
  • A. 2:1:4
  • B. 4:1:4
  • C. 1:2:4
  • D. 1:1:4

Solution

Core Logic

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

a = (1 - 2(1) + 2(1)²)²⁰²³ (3 - 4(1)² + 2(1)³)²⁰²⁴ a = (1)²⁰²³ (1)²⁰²⁴ = 1
Step 1: Evaluate Limit for b

Evaluate b = x → 0 ∫₀x ln(1 + t)1 + t²⁰²⁴ dtx² Using L'Hôpital's Rule (differentiating numerator via Newton-Leibniz):

b = x → 0 ln(1 + x)1 + x²⁰²⁴2x = x → 0 (ln(1 + x))/(x) × 12(1 + x²⁰²⁴) b = 1 × (1)/(2) = (1)/(2)
Step 2: Analyze Common Roots

The given second equation is 2bx² + ax + 4 = 0. Substitute a = 1 and b = (1)/(2):

2((1)/(2))x² + 1(x) + 4 = 0 x² + x + 4 = 0

The discriminant of x² + x + 4 = 0 is D = 1 - 16 < 0. Roots are non-real complex conjugates.

Step 3: Final Ratio

Since c, d, e in R and one root is common with a quadratic having non-real roots, both roots must be common. Thus, the coefficients must be proportional:

(c)/(1) = (d)/(1) = (e)/(4)

This implies d : c : e = 1 : 1 : 4.

Pattern Recognition

If a quadratic equation with real coefficients shares a common root with another quadratic having complex roots (D < 0), both roots must be shared, meaning their coefficients are directly proportional.

Chapter Mix

Class 11 Maths: Binomial Theorem Class 12 Maths: Limits and Derivatives Class 11 Maths: Quadratic Equations

Q25 jee_main_2024_31_jan_morning Coefficients in Expansion
In the expansion of (1 + x)(1 - x²)(1 + (3)/(x) + (3)/(x²) + (1)/(x³))⁵, x ≠ 0, the sum of the coefficient of x³ and x⁻¹³ is equal to
Numerical Answer. Answer: 118 to 118

Solution

Core Logic
(1+x)(1-x²) ( (1 + (1)/(x))³ )⁵ = (1+x)(1-x)(1+x) (x+1)¹⁵x¹⁵ = (1-x)(1+x)¹⁷x¹⁵ = (1+x)¹⁷ - x(1+x)¹⁷x¹⁵
Step 1: Find Coefficient of x^3

To find coeff of x³ in (1+x)¹⁷ - x(1+x)¹⁷x¹⁵, we need the coeff of x¹⁸ in the numerator (1+x)¹⁷ - x(1+x)¹⁷. The maximum power of x in (1+x)¹⁷ is 17, and in x(1+x)¹⁷ is 18. Coeff of x¹⁸ in (1+x)¹⁷ is 0. Coeff of x¹⁸ in x(1+x)¹⁷ is the coeff of x¹⁷ in (1+x)¹⁷, which is 1717 = 1. Thus, coeff of x¹⁸ in the numerator is 0 - 1 = -1.

Step 2: Find Coefficient of x^{-13}

To find coeff of x⁻¹³, we need the coeff of x² in the numerator (1+x)¹⁷ - x(1+x)¹⁷. Coeff of x² in (1+x)¹⁷ is 172. Coeff of x² in x(1+x)¹⁷ is coeff of x¹ in (1+x)¹⁷, which is 171. Value = 172 - 171 = (17 × 16)/(2) - 17 = 136 - 17 = 119.

Step 3: Final Sum

Sum of coefficients = -1 + 119 = 118.

Chapter Mix

Class 11 Maths: Binomial Theorem

More Binomial Theorem Questions — jee_main_2025_03_april_morning

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