Problem 28 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If for positive integers \(r>1, n>2\), the coefficients of the \((3 r)^{t h}\) and \((r+2)^{\text {th }}\) powers of \(x\) in the expansion of \((1+x)^{2 n}\) are equal, then \(n\) is equal to: (a) \(2 r+1\) (b) \(2 r-1\) (c) \(3 r\) (d) \(r+1\)

Step-by-Step Solution

Answer

The correct answer is (a) \( n = 2r + 1 \).

Step 1: Understand the coefficients in binomial expansion

In the binomial expansion of \( (1+x)^{2n} \), the coefficient of \( x^k \) is given by \( \binom{2n}{k} \). We need to find the conditions where the coefficients of the \( (3r)^{th} \) and \( (r+2)^{th} \) powers of \( x \) are equal.

Step 2: Set up the equality of coefficients

According to the problem, \( \binom{2n}{3r} = \binom{2n}{r+2} \). This equation arises from setting the coefficients of the \( (3r)^{th} \) term and the \( (r+2)^{th} \) term in the expansion equal.

Step 3: Simplify using binomial coefficient symmetry

Using the property of binomial coefficients, \( \binom{n}{k} = \binom{n}{n-k} \), we know \( \binom{2n}{3r} = \binom{2n}{2n-3r} \). Combine this with our equation to set up a relation: \( 3r = 2n-(r+2) \).

Step 4: Solve the equation for n

Rearranging \( 3r = 2n - r - 2 \) gives \( 2n = 4r + 2 \). Solving for \( n \), we divide each side by 2: \( n = 2r + 1 \).

Step 5: Verify the solution satisfies conditions

Check that when \( n = 2r + 1 \), all conditions in the given problem such as \( r > 1 \) and \( n > 2 \) are met. Substituting any integer \( r > 1 \) provides a valid integer \( n > 2 \).

Key Concepts

Binomial Coefficients

The concept of binomial coefficients is crucial in understanding the binomial theorem. A binomial coefficient is represented as \( \binom{n}{k} \) and read as "n choose k". It gives the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to the order of selection.
This coefficient can be calculated using the formula:

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

Where \( n! \) denotes the factorial of \( n \). Factorials are the products of all positive integers up to \( n \). Binomial coefficients are central in the expansion of expressions like \((1+x)^n\) and play a key role in various fields of mathematics.

Positive Integers

Positive integers are the set of all whole numbers greater than zero. They include numbers like 1, 2, 3, and so on. Positive integers are fundamental in various mathematical concepts, including the problem at hand, which involves different integer values of \( r \) and \( n \). In the exercise, we specify conditions such as \( r>1 \) and \( n>2 \), which implies that both \( r \) and \( n \) must be positive, thus ensuring that they satisfy the requirements of being whole numbers distinctly larger than zero.
Positive integers are crucial as they form the basis for operations in number theory and are inherently involved in computations involving the binomial theorem and polynomial expansions.

Symmetry Property

In the realm of binomial coefficients, the symmetry property is an elegant and helpful feature. This property states that \( \binom{n}{k} = \binom{n}{n-k} \). This means that choosing \( k \) objects from \( n \) is the same as leaving \( n-k \) objects, resulting in the same number of combinations.
This property is handy when simplifying expressions and solving equations involving binomial coefficients because it often reduces the complexity of calculations.
In our specific problem, we use this property to equate \( \binom{2n}{3r} = \binom{2n}{r+2} \) by rewriting \( \binom{2n}{3r} \) as \( \binom{2n}{2n-3r} \), allowing us to simplify and solve the equation for \( n \).

Polynomial Expansion

Polynomial expansion is a method where you express a binomial raised to a power as a sum of terms involving binomial coefficients and products of the variables. The binomial theorem provides a way to expand expressions of the form \((1+x)^n\).
Every term in this expansion can be expressed as \( \binom{n}{k} x^k \), where \( k \) varies from 0 to \( n \). This gives each term a specific coefficient known as a binomial coefficient.
In the given exercise, polynomial expansion is used to identify terms that contain specific powers of \( x \) and their corresponding coefficients. Solving the problem involves equating these coefficients to find relationships between different powers, which ultimately helps determine the unknown values, such as \( n \) in this case.
Understanding polynomial expansion is vital for tackling many algebraic problems and is frequently applied in calculus, numerical analysis, and discrete mathematics.