Binomial Theorem & Expansions
Ace JEE Main Binomial Theorem with authentic PYQ speed drills. Master general term T_(r+1), term independent of x, divisibility & coefficient sum tricks.
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4-Step Solution & Tricks
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Binomial Theorem & Expansions:
- General Term T_(r+1) = ⁿC_r aⁿ⁻ʳ bʳ
- Term Independent of x & Constant Terms
- Sum of Binomial Coefficients & Alternating Sums
- Divisibility & Remainder Problems using Binomial Expansion
- Greatest Term & Multinomial Theorem
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Binomial Theorem questions in under 45 seconds:
Sum of coefficients: Simply substitute x = 1 in the given expansion.
Remainder of (a + 1)ⁿ divided by a: Always 1 for positive integers.
ⁿC_r + ⁿC_(r-1) = ⁿ⁺¹C_r (Pascal’s Rule).
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: -144
- Step 1: Factor the polynomial: 1 - x - x² + x³ = (1 - x)(1 - x²).
- Step 2: Raise to power 6: [(1 - x)(1 - x²)]⁶ = (1 - x)⁶ (1 - x²)⁶.
- Step 3: Expand both terms using Binomial theorem: ∑ ⁶Cᵣ (-x)ʳ · ∑ ⁶Cₖ (-x²)ᵏ.
- Step 4: For power x⁷, r + 2k = 7. Evaluating valid integer pairs gives the coefficient -144.
⚡ Speed Tip: Factorization first: (1 - x - x² + x³) = (1 - x)(1 - x²). Always factor before expanding.
💡 View Step-by-Step Solution & Answer
Correct Answer: 7/18
- Step 1: General term is T_{r+1} = ⁹Cᵣ (3x²/2)⁹⁻ʳ (-1/(3x))ʳ.
- Step 2: Power of x is 2(9 - r) - r = 18 - 3r.
- Step 3: Set power to 0: 18 - 3r = 0 ⇒ r = 6.
- Step 4: T₇ = ⁹C₆ (3/2)³ (-1/3)⁶ = 84 · (27/8) · (1/729) = 7/18.
⚡ Speed Tip: Shortcut: r = (n · p) / (p + q) = (9 × 2)/(2 + 1) = 6.
💡 View Step-by-Step Solution & Answer
Correct Answer: 1
- Step 1: 2024 = 7 × 289 + 1 ⇒ 2024 ≡ 1 (mod 7).
- Step 2: (2024)²⁰²⁴ ≡ 1²⁰²⁴ (mod 7).
- Step 3: 1²⁰²⁴ = 1.
- Step 4: The remainder is 1.
⚡ Speed Tip: Modular Arithmetic: 2024 ≡ 1 (mod 7). (1)ⁿ = 1.
❓ Frequently Asked Questions (FAQs)
How to find remainder of large powers in JEE Main?
Write the base in form (k·m ± 1)ⁿ where m is the divisor, then expand binomially to leave only the last term.