Problem 56 - A Complete Resource Book in Mathematics for JEE Main

Question

Given positive integers \(r>1, n>2\) and the coefficients of \((3 r)\) th term and \((r+2)\) th term in the binomial expansion of \((1+x)^{2 \mathrm{n}}\) are equal, then \(r=\) (A) \(\frac{n}{2}, n\) even (B) \(\frac{n}{2}\) (C) \(n\) (D) 1

Step-by-Step Solution

Answer

Option (B), \(r = \frac{n}{2}\).

Step 1: Understand the Binomial Coefficient

In a binomial expansion \((1+x)^{2n}\), the general term is given by \(T_{k+1} = \binom{2n}{k}x^k\). Here, \(\binom{2n}{k}\) denotes the binomial coefficient which is calculated as \(\frac{(2n)!}{k!(2n-k)!}\).

Step 2: Equate the Given Terms

According to the problem, the coefficients of the \((3r)\)th term and the \((r+2)\)th term are equal. Therefore, we have the equation: \(\binom{2n}{3r-1} = \binom{2n}{r+1}\).

Step 3: Simplify the Binomial Coefficient Equation

Using the property that \(\binom{n}{k} = \binom{n}{n-k}\), we can rewrite the equation as \(\binom{2n}{3r-1} = \binom{2n}{r+1}\) implies \(3r - 1 = 2n - (r + 1)\).

Step 4: Solve for r

Simplify the equation obtained in Step 3: \[3r - 1 = 2n - r - 1\]\[3r = 2n - r\]Add \(r\) to both sides:\[4r = 2n\]Divide both sides by 4:\[r = \frac{n}{2}\].

Step 5: Check Additional Condition

Verify that \(r = \frac{n}{2}\) satisfies the condition of being greater than 1. This implies \(n > 2\), so \(\frac{n}{2} > 1\). Thus, the solution is consistent with the initial conditions that \(r > 1\).

Key Concepts

Understanding the Binomial Coefficient

In the binomial expansion \((1+x)^{2n}\), we come across something called the "binomial coefficient," which plays a crucial role.The binomial coefficient counts how many ways we can choose a subset of items from a larger set, without considering the order.Mathematically, it is represented in the form \(\binom{2n}{k}\),where \(2n\) is the total number of items, and \(k\) is the number of items we want to choose.

The formula to calculate it is:\[\binom{2n}{k} = \frac{(2n)!}{k!(2n-k)!}\]Here, \(!\) means factorial, which is the product of an integer and all the integers below it.For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1\).
The property of binomial coefficients, such as \(\binom{n}{k} = \binom{n}{n-k}\), comes in handy.This property helps in simplifying equations when solving problems.

In this exercise, this property was used to equate terms and finally solve for \(r\).

Applying Mathematical Problem-Solving

The key to solving a mathematical problem is to understand the given data and the relationships between them.In our exercise, we began by examining the coefficients of specific terms in a binomial expansion.These terms were "\((3r)\)th term"and "\((r+2)\)th term."After stating the problem, the next step was to translate "coefficients are equal" into a mathematical equation: \(\binom{2n}{3r-1} = \binom{2n}{r+1}\).

Problem-solving involves organizing the information,making assumptions if necessary,and rigorously following logical steps to find a solution.In this exercise, we exploited symmetry properties of the binomial coefficient to simplify the problem.By rewriting the equality and eventually using simple algebra we derived a usable equation from which \(r\) can be extracted as \( r = \frac{n}{2}\).

Deciphering Algebraic Expressions

Algebraic expressions form the backbone of mathematical problem-solving, allowing us to describe mathematical ideas concisely. In the given problem, our algebraic expressions consist of terms from a binomial expansion. The terms themselves are made up of binomial coefficients and power terms, showing the dependence of each term’s value on the chosen index and variable.

Underpinning our work with these expressions involves understanding properties, like how substituting or rearranging terms can lead to revealing new insights. The step-by-step analysis in this problem employed algebraic skills such as:

• Expanding expressions
• Simplifying equations
• Recognizing symmetrical properties

These skills, when applied correctly, reduce a complex-sounding problem into simpler, hence more manageable, parts. Effectively working with algebraic expressions is like cracking a code, where each line of work provides further insight to unlock the overall solution.