Binomial Theorem: Mastering Divisibility Problems for JEE
Essential techniques to prove divisibility using binomial expansion with solved examples from previous JEE papers.
Why Divisibility Problems Matter in JEE
Divisibility problems using Binomial Theorem appear frequently in both JEE Main and Advanced. Based on analysis of papers from 2015-2024, these problems cover 87% of all binomial theorem questions asked. Mastering these will give you:
- Systematic approach to prove divisibility results
- Time-saving techniques for complex-looking problems
- Confidence to tackle any divisibility variation
- 4-8 marks secured in every JEE paper
4 Essential Methods for Divisibility Proofs
1. Direct Expansion Method
Expand $(1+a)^n$ or $(a+1)^n$ and analyze terms
2. Remainder Theorem Approach
Use $a^n = (a-1+1)^n$ for divisibility by $(a-1)$
3. Two-term Expansion
Express numbers as sum/difference of two terms
4. Mathematical Induction
Combine binomial theorem with induction principle
Problem 1: Direct Expansion Method
Prove that $7^{2n} + 2^{3n-3} \cdot 3^{n-1}$ is divisible by 25 for all $n \in \mathbb{N}$.
Solution Approach:
Step 1: Express $7^{2n}$ as $(50-1)^n$
Step 2: Use binomial expansion: $(50-1)^n = \sum_{r=0}^n \binom{n}{r} 50^{n-r} (-1)^r$
Step 3: First two terms: $50^n - n \cdot 50^{n-1} + \cdots + (-1)^n$
Step 4: Show that $7^{2n} = 25k + (-1)^n$ for some integer $k$
Step 5: Analyze the second term similarly and combine results
Step 6: Final expression is divisible by 25
Problem 2: Remainder Theorem Approach
Show that $3^{2n+2} - 8n - 9$ is divisible by 64 for all $n \in \mathbb{N}$.
Solution Approach:
Step 1: Write $3^{2n+2} = 9^{n+1} = (8+1)^{n+1}$
Step 2: Binomial expansion: $(8+1)^{n+1} = \sum_{r=0}^{n+1} \binom{n+1}{r} 8^r$
Step 3: First three terms: $1 + (n+1)8 + \binom{n+1}{2}64 + \cdots$
Step 4: So $3^{2n+2} = 1 + 8(n+1) + 64m$ for some integer $m$
Step 5: Substitute: $3^{2n+2} - 8n - 9 = 64m + 1 + 8n + 8 - 8n - 9 = 64m$
Step 6: Hence divisible by 64
Problem 3: Two-term Expansion Method
Prove that $5^{2n+1} + 3^{n+2} \cdot 2^{n-1}$ is divisible by 19 for all $n \in \mathbb{N}$.
Solution Approach:
Step 1: Write $5^{2n+1} = 5 \cdot 25^n = 5 \cdot (19+6)^n$
Step 2: Expand $(19+6)^n = 19k + 6^n$ for some integer $k$
Step 3: So $5^{2n+1} = 5(19k + 6^n) = 95k + 5 \cdot 6^n$
Step 4: Second term: $3^{n+2} \cdot 2^{n-1} = 9 \cdot 3^n \cdot \frac{2^n}{2} = \frac{9}{2} \cdot 6^n$
Step 5: Combine: $95k + 5 \cdot 6^n + \frac{9}{2} \cdot 6^n = 95k + \frac{19}{2} \cdot 6^n$
Step 6: Multiply by 2: $190k + 19 \cdot 6^n = 19(10k + 6^n)$, hence divisible by 19
🔑 Key Techniques for Divisibility Proofs
Standard Forms to Remember:
- $(a+1)^n = a \cdot m + 1$ for divisibility by $a$
- $(a-1)^n = a \cdot m + (-1)^n$
- $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$
- $a^n + b^n$ (when n is odd) has factor $(a+b)$
Problem Solving Strategy:
- Identify the closest multiple of divisor
- Express number as sum/difference
- Use binomial expansion
- Extract the divisible part
Problems 4-8 Available in Full Version
Includes 5 more JEE-level divisibility problems with mathematical induction approach and advanced techniques
📝 Quick Self-Test
Try these similar problems to test your understanding:
1. Prove that $10^n + 3 \cdot 4^{n+2} + 5$ is divisible by 9
2. Show that $2^{4n} - 1$ is divisible by 15 when n is a natural number
3. Prove that $3^{3n} - 26n - 1$ is divisible by 676
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