Problem 56 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \((1+x)^{2 n}\) are equal, then \(r=\) (A) \(\frac{n}{2}, n\) even (B) \(\frac{n}{2}\) (C) \(n\) (D) 1

Step-by-Step Solution

Answer

\(r = \frac{n}{2}\), \(n\) is even (Option A).

Step 1: Understanding Binomial Coefficient

In the binomial expansion of (1+x)^{2n}), the j-th term can be expressed as \(\binom{2n}{j-1}x^{j-1}\). Therefore, the coefficient of the \((3r)\)th term is \(\binom{2n}{3r-1}\) and the coefficient of the \((r+2)\)th term is \(\binom{2n}{r+1}\).

Step 2: Setting Up the Equality

Given that these two coefficients are equal, we set up the equation:\[\binom{2n}{3r-1} = \binom{2n}{r+1}\] This equality implies symmetry in binomial coefficients, which occurs when the indices of the coefficients sum to the power of the binomial, i.e., the indices \((3r-1)\) and \((r+1)\) must add up to \((2n+1)\).

Step 3: Solving for r

Using the symmetry property, we set the equation:\[(3r-1) + (r+1) = 2n\]Simplifying, this gives:\[4r = 2n\]\[r = \frac{n}{2}\]

Step 4: Conclusion and Validity Check

Since \(n > 2\) and must be a positive integer, \(\frac{n}{2}\) should also be an integer, indicating \(n\) is even. This matches option (A).

Key Concepts

Binomial Coefficients

Binomial coefficients are a key part of the Binomial Theorem, used to expand polynomials raised to a power.
In mathematical terms, for a binomial \((1+x)^n\), the binomial coefficients are denoted as \(\binom{n}{k}\), where \(n\) is the power of the polynomial, and \(k\) is the specific term.
These coefficients represent the number of ways to choose \(k\) elements from \(n\) elements without regard to order.

• They can be calculated using the formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).
• Binomial coefficients are symmetric, meaning \(\binom{n}{k} = \binom{n}{n-k}\).

In the context of the original exercise, understanding these coefficients allows us to identify equal terms in the expansion, as discussed in the solution.

Symmetry in Expansion

Symmetry in binomial expansion is a fascinating property which states that the coefficients of terms equidistant from either end of the expansion are equal.
This property is deeply connected to the inherent symmetry of binomial coefficients.

• For instance, in an expansion of \((1+x)^{2n}\), the coefficient of the \(k\)-th term is equal to the coefficient of the \((2n-k+1)\)-th term.
• This symmetry arises because both terms have the same number of elements chosen but from "opposite ends" of the sequence.

In the given exercise, this concept is used to say that for the coefficients to be equal, the indices of the terms \((3r-1)\) and \((r+1)\) must sum to \(2n\), showcasing the beauty of symmetry in binomials.

Positive Integers in Binomials

Understanding the role of positive integers in binomials is essential, especially in terms of how they influence the terms of the expansion.
In the binomial expansion, positive integers represent the power to which the binomial is raised, affecting both the number of terms and the possible values for coefficients.

• In the polynomial \((1+x)^{2n}\), \(n\) is a positive integer determining the number of terms, which is \(2n+1\).
• Each term in the expansion is defined by an integer index \(k\), and its position and value depend on this index.

For problems like the given exercise, knowing that \(r\) and \(n\) are positive integers enhances our ability to manipulate and solve the expressions accurately, as shown in the equality \[4r = 2n\] from the solution, ultimately using these properties to arrive at \(r = \frac{n}{2}\).