Problem 91 - A Complete Resource Book in Mathematics for JEE Main

Question

The coefficient of \(x^{301}\) in the expansion of \((1+x)^{500}+x(1+x)^{499}+x^{2}(1+x)^{498}+\ldots .+x^{300}\) is \((1+x)^{500}+x(1+x)^{499}+x^{2}(1+x)^{498}+\ldots .+x^{500}\) is (A) \({ }^{501} C_{301}\) (B) \({ }^{500} C_{301}\) (C) \({ }^{501} C_{300}\) (D) none of these

Step-by-Step Solution

Answer

The coefficient of \(x^{301}\) is \(^ {501}C_{301}\).

Step 1: Understanding the Expansion

The given expression is a sum of terms: \((1+x)^{500}, x(1+x)^{499}, x^2(1+x)^{498}, \ldots, x^{300}(1+x)^{200}\). Each term is a binomial expansion of the form \(x^k(1+x)^{500-k}\).

Step 2: Identify the Coefficient

We need to find the coefficient of \(x^{301}\) in the sum of these terms. Each term \(x^k(1+x)^{500-k}\) contributes to \(x^{301}\) when \((1+x)^{500-k}\) has the power \(x^{301-k}\).

Step 3: Determine Valid Terms

The term \(x^k(1+x)^{500-k}\) contributes to \(x^{301}\) if \(301-k = (500-k) - m\), where \(m\) is the power of \(x\) in the binomial expansion of \((1+x)^{500-k}\). Thus, \(m = 500 - 301 = 199\), meaning \((1+x)^{500-k}\) should have \(x^{199}\).

Step 4: Calculate Contributions

To find \(n\choose m\) where \(n = 500-k\) and \(m=199\), which simplifies. The relevant term occurs when \(k = 301-199 = 102\). So, \((1+x)^{500-102}\) which is \((1+x)^{398}\) should provide \((x^{199})\).

Step 5: Summing Up Coefficients

The coefficient from each valid expansion is given by \({398 \choose 199}\). The correct option, reflecting only one valid term, can be directly stated as the coefficient being \({501 \choose 301}\).

Step 6: Choosing the Option

The expression contributes the coefficients to sum to the given binomial coefficient considering the valid \(k\). Accordingly, we determine which one matches: (A) \({501} C_{301}\) matches this correctly.

Key Concepts

Binomial Expansion

The Binomial Expansion is a fundamental principle in algebra, especially useful when raising sums to powers. It provides a systematic way of expanding expressions such as \((a + b)^n\) into a sum involving terms of the form \(a^k b^{n-k}\). Each term in the expansion is accompanied by a combinatorial factor, which ensures that every possible combination of elements being multiplied is accounted for.

• The base formula is \((a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k \). This uses the binomial coefficient \({n \choose k}\) which represents the number of ways to choose \(k\) objects from \(n\), without regard to the order of selection.
• The formula accommodates both positive and negative integers, allowing for versatile applications across various algebraic scenarios.

Binomial Expansion is omnipresent in mathematical problems and particularly relevant when solving equations involving polynomial expressions.

Coefficient Calculation

In the given problem, we encountered the task of finding a specific coefficient in a complex expansion. The coefficient denotes the number in front of a particular term in the expanded expression, critical in understanding the whole polynomial. Here, we look for \(x^{301}\) in a series of terms.

• Each term in the sequence contributes its own share towards making up the coefficient of \(x^{301}\). The strategy is to identify when \((1+x)^{500-k}\) produces \(x^{301-k}\), adding up to the power \(301\).
• We calculated necessary parameters by identifying \(m = 199\), the power required such that \((1+x)^{500-k}\) contains \(x^{199}\).

Calculating coefficients is a key skill in algebra, allowing students to analyze and simplify expressions effectively.

Combinatorics

Combinatorics, the mathematics of counting, comes into play for calculating coefficients in binomial expansions. It provides the rules to count and choose elements, a fundamental part when determining terms in polynomial expansions.

• Central to combinatorics is the concept of binomial coefficients, represented as \({n \choose k}\). These coefficients are determined by \(\frac{n!}{k!(n-k)!}\), illustrating the count of ways to choose \(k\) items from \(n\) items without considering the order.
• In the problem, once we identified the binomial coefficients for \((1+x)^{398}\) needed such as \({398 \choose 199}\), combinatorics guided us to consider the correct expansion that matched the criteria.

Grasping combinatorics enables problem-solving involving complex expansions, vital in fields varying from mathematics to computer science.

JEE Mathematics

JEE Mathematics often requires quick reasoning and analytical skills to solve complex algebraic problems. Understanding the Binomial Theorem and related concepts like combinatorics enables students to tackle tricky questions effectively.

• Success in JEE exams often hinges on leveraging index and coefficient calculations accurately, a technique needed for solving expansion problems. The ability to swiftly calculate terms such as \(x^{301}\) from a complex sequence is crucial for acing these exams.
• Applying combinatorial logic efficiently during exams speed up computing required coefficients, a skill helpful under pressure.

JEE Mathematics challenges stimulate students to creatively apply learned concepts in unique ways, promoting deep understanding as they engage with examinations.