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Binomial Theorem appeared 37 times across 3 years — 4.3% of Mathematics. This question is from Rational Terms in Expansion.

Year 2026 2025 2024 Total
Questions 9 17 11 37

The sum of all rational terms in the expansion of (2 + √(3))⁸ is:

Solution & Explanation

Related Formula

General term in the binomial expansion of (a+b)ⁿ:

Tᵣ₊₁ = ⁿCᵣ · an-r · b^r
Core Logic

For a term to be rational in (2+√(3))⁸, the power of √(3) must be an even integer[cite: 1218]. Thus, r can only take even values from 0 to 8

r in 0, 2, 4, 6, 8
Step 1: Calculating individual rational terms

Sum up terms explicitly for all valid values of r :

Sum = ⁸C₀(2)⁸ + ⁸C₂(2)⁶(√(3))² + ⁸C₄(2)⁴(√(3))⁴ + ⁸C₆(2)²(√(3))⁶ + ⁸C₈(√(3))⁸

Compute the individual numeric values :

  • r=0: 1 · 256 = 256
  • r=2: 28 · 64 · 3 = 5376
  • r=4: 70 · 16 · 9 = 10080
  • r=28 · 4 · 27 = 3024
  • r=8: 1 · 1 · 81 = 81
Total Sum = 256 + 5376 + 10080 + 3024 + 81 = 18817
Pattern Recognition

Notice that the sum of all rational terms can also be viewed as the rational part of the expanded configuration, equivalent to (2+√(3))⁸ + (2-√(3))⁸2.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

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

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