Problem 136 - A Complete Resource Book in Mathematics for JEE Main

Question

The coefficient of \(x^{7}\) in the expansion of the expression \(\left(1-x-x^{2}+x^{3}\right)^{6}\) is (A) \(-132\) (B) \(-144\) (c) 132 (D) 144

Step-by-Step Solution

Answer

The coefficient of \(x^7\) is \(-132\) (Option A).

Step 1: Identify the problem

We are asked to find the coefficient of \(x^7\) in the expansion of the expression \((1-x-x^2+x^3)^6\).

Step 2: Recognize the usefulness of the multinomial theorem

To find the coefficient of a specific term in a multinomial expansion like \((a+b+c+d)^n\), the multinomial theorem is handy. Here, each term can be treated as different components of the multinomial to find the desired power of \(x\).

Step 3: Establish the multinomial formula

The coefficient of \(x^7\) can be derived using the multinomial expansion:\[ (1-x-x^2+x^3)^6 = \sum_{i+j+k+l=6} \frac{6!}{i!j!k!l!} \cdot (-1)^j \cdot (-1)^k \cdot x^{j+2k+3l} \]where \(i+j+k+l=6\) and \(j+2k+3l=7\).

Step 4: Solve for specific combinations

Find combinations of \(i, j, k, l\) such that both \(i+j+k+l=6\) and \(j+2k+3l=7\):- \((i, j, k, l) = (3, 2, 1, 0)\)- \((i, j, k, l) = (0, 1, 2, 3)\) and analyze the possibility of others.

Step 5: Calculate the coefficients of valid combinations

For \((i, j, k, l) = (3, 2, 1, 0)\), the term contributes:- Coefficient: \( \frac{6!}{3!2!1!0!} \times (-1)^2 \times (-1)^1 = 60 \)For \((i, j, k, l) = (0, 1, 2, 3)\), the term contributes:- Coefficient: \( \frac{6!}{0!1!2!3!} \times (-1)^1 \times (-1)^2 = -192 \)

Step 6: Combine coefficients to find the answer

Add the coefficients from Step 5:\(60 - 192 = -132\).Thus, the coefficient of \(x^7\) is \(-132\).

Key Concepts

Binomial Theorem

The binomial theorem is a fundamental concept in mathematics that describes the algebraic expansion of powers of a binomial expression. In its simplest form, the theorem uses binomial coefficients to expand expressions like

• \((a + b)^n\)

into a sum involving terms of the form

• \(a^k b^{n-k}\),

where each term is multiplied by a binomial coefficient. These coefficients are given by

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

In the context of our original exercise, while it deals with a multinomial rather than a simple binomial, understanding the binomial theorem helps develop the base idea of arranging terms and applying coefficients in polynomial expansions.
It's important to recognize that the binomial theorem is a special case of the broader concept of multinomial expansion, a more complex formulation that takes into account more than two components in the base expression.

Polynomial Coefficients

Polynomial coefficients appear in multiple areas of algebra, often representing constants attached to variables of different powers. When considering expansions, like in our original problem, identifying the correct polynomial coefficient involves calculating these constants using a structured formula.

The task is to figure out how each term in the expanded expression contributes to the final sum, represented by a specific power, such as\(x^7\) in the original exercise. In multinomial expansions, like our example

• \((1-x-x^2+x^3)^6\),

coefficients are derived through combinations where respective powers of each term sum to the desired power. Specifically, our solution demonstrates combining calculated coefficients of valid terms like:

• \(\frac{6!}{3!2!1!0!}\) for one configuration,
• \(\frac{6!}{0!1!2!3!}\) for another,

then using properties of negative signs to get the actual coefficients. Paying close attention to these details allows the determination of accurate polynomial coefficients for targeted variables.

Combinatorics in Algebra

Combinatorics plays a crucial role in algebra, often intersecting with concepts such as polynomial expansions and coefficients. In any expansion problem, selections of terms or coefficients form combinations to address conditions set by the problem statement.

In the example of

• \((1-x-x^2+x^3)^6\),

our goal involved using Combinatorics to find suitable combinations of our expression variables

• \((i,j,k,l)\)

that satisfy two conditions: \(i+j+k+l=6\) and \(j+2k+3l=7\). With these valid configurations identified, mathematical calculations align with combinatorial logic to generate each term's coefficient.

Combinatorial practice involves systematic problem-solving, moving from idea conception to conclusion by organizing and calculating these combinations. Such techniques empower us to solve complex expressions efficiently, reinforcing algebra's structure while dissecting polynomial challenges like the one outlined in the original exercise.