Binomial Theorem: Handling Complex Expansions - $(x + y + z)^n$
Master the multinomial theorem, coefficient problems, and advanced expansion techniques for JEE success.
Why Complex Binomial Expansions Matter
Expansions of $(x+y+z)^n$ and similar expressions appear frequently in JEE Main and Advanced. Based on analysis of past papers, these problems test:
- Understanding of multinomial theorem and its applications
- Coefficient extraction techniques for specific terms
- Problem-solving with constraints on exponents
- Algebraic manipulation skills with multiple variables
The Multinomial Theorem
For $(x_1 + x_2 + \cdots + x_m)^n$, the expansion is given by:
$$(x_1 + x_2 + \cdots + x_m)^n = \sum \frac{n!}{n_1! n_2! \cdots n_m!} x_1^{n_1} x_2^{n_2} \cdots x_m^{n_m}$$
where the summation is over all non-negative integers $n_1, n_2, \ldots, n_m$ such that $n_1 + n_2 + \cdots + n_m = n$.
Problem 1: Basic Multinomial Expansion
Find the coefficient of $x^2 y^3 z^2$ in the expansion of $(x + y + z)^7$.
Solution Approach:
Step 1: Using multinomial theorem, the general term is:
$$T = \frac{7!}{a! b! c!} x^a y^b z^c$$
where $a + b + c = 7$
Step 2: We need $a = 2$, $b = 3$, $c = 2$ (satisfying $2+3+2=7$)
Step 3: Coefficient = $\frac{7!}{2! 3! 2!} = \frac{5040}{2 \cdot 6 \cdot 2} = \frac{5040}{24} = 210$
Step 4: Final answer: $\boxed{210}$
Problem 2: Coefficient with Constraints
Find the number of distinct terms in the expansion of $(1 + x + x^2 + x^3)^5$.
Solution Approach:
Step 1: The expression is $(1 + x + x^2 + x^3)^5$
Step 2: General term: $T = \frac{5!}{a! b! c! d!} 1^a x^b (x^2)^c (x^3)^d$
where $a + b + c + d = 5$
Step 3: Power of $x$ in the term = $b + 2c + 3d$
Step 4: Minimum power: when $b=c=d=0$, power = 0
Step 5: Maximum power: when $a=b=c=0, d=5$, power = 15
Step 6: All integer powers from 0 to 15 are possible
Step 7: Number of distinct terms = $15 - 0 + 1 = 16$
Step 8: Final answer: $\boxed{16}$
Problem 3: Specific Term with Conditions
Find the coefficient of $x^5$ in the expansion of $(1 + x + x^2 + x^3)^6$.
Solution Approach:
Step 1: We need to find all non-negative integer solutions to:
$a + b + c + d = 6$ and $b + 2c + 3d = 5$
Step 2: Case analysis:
Case 1: $d = 0$, then $b + 2c = 5$ and $a + b + c = 6$
• $c = 0$: $b = 5$, then $a = 1$ → coefficient = $\frac{6!}{1!5!0!0!} = 6$
• $c = 1$: $b = 3$, then $a = 2$ → coefficient = $\frac{6!}{2!3!1!0!} = 60$
• $c = 2$: $b = 1$, then $a = 3$ → coefficient = $\frac{6!}{3!1!2!0!} = 60$
Case 2: $d = 1$, then $b + 2c = 2$ and $a + b + c = 5$
• $c = 0$: $b = 2$, then $a = 3$ → coefficient = $\frac{6!}{3!2!0!1!} = 60$
• $c = 1$: $b = 0$, then $a = 4$ → coefficient = $\frac{6!}{4!0!1!1!} = 30$
Case 3: $d = 2$, impossible since $3d = 6 > 5$
Step 3: Total coefficient = $6 + 60 + 60 + 60 + 30 = 216$
Step 4: Final answer: $\boxed{216}$
🚀 Problem-Solving Strategies
For Multinomial Expansions:
- Always verify $n_1 + n_2 + \cdots + n_m = n$
- Use the formula $\frac{n!}{n_1! n_2! \cdots n_m!}$ for coefficients
- For specific powers, set up equations for exponents
- Use case analysis when constraints are involved
For Coefficient Problems:
- Convert to simpler forms using substitutions
- Use combinatorial identities when possible
- Look for patterns in the exponents
- Check if the problem can be solved using generating functions
Problems 4-8 Available in Full Version
Includes 5 more complex binomial expansion problems with detailed solutions:
- Greatest coefficient problems
- Expansions with negative exponents
- Applications in probability
- Problem-solving with inequalities
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Find the coefficient of $x^3y^2z^4$ in $(x+y+z)^9$
2. Find the number of terms in $(1+x+x^2)^8$
3. Find the sum of all coefficients in $(x+2y+3z)^5$
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