Problem 26 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\left(2+\frac{x}{3}\right)^{55}\) is expanded in the ascending powers of \(x\) and the coefficients of powers of \(x\) in two consecutive terms of the expansion are equal, then these terms are: (a) \(7^{\text {th }}\) and \(8^{\text {th }}\) (b) \(8^{\text {th }}\) and \(9^{\text {th }}\) (c) \(28^{\text {th }}\) and \(29^{\text {th }}\) (d) \(27^{\text {th }}\) and \(28^{\text {th }}\)

Step-by-Step Solution

Answer

(c) 28th and 29th

Step 1: Understand the Question

We need to find two consecutive terms in the binomial expansion of \(\left(2+\frac{x}{3}\right)^{55}\) whose coefficients of \(x\) powers are equal. The terms in the expansion are given by the binomial theorem.

Step 2: Apply the Binomial Theorem

The \(r\)-th term of \(\left(2+\frac{x}{3}\right)^{55}\) is given by \(T_r = \binom{55}{r-1} \cdot 2^{55-(r-1)} \cdot \left(\frac{x}{3}\right)^{r-1}\). This can be simplified to \(T_r = \binom{55}{r-1} \cdot 2^{56-r} \cdot \frac{x^{r-1}}{3^{r-1}}\).

Step 3: Set Consecutive Coefficients Equal

To find consecutive terms with equal coefficients, set the coefficients of \(x^{r-1}\) and \(x^r\) in the \(r\)th and \((r+1)\)th terms equal: \(\binom{55}{r-1} \cdot 2^{56-r} \cdot \frac{1}{3^{r-1}} = \binom{55}{r} \cdot 2^{55-r} \cdot \frac{1}{3^r}\).

Step 4: Simplify and Solve for r

Simplifying the equation gives \(\frac{\binom{55}{r-1}}{\binom{55}{r}} = \frac{2}{3}\). Using the binomial coefficient identity and simplifying, we get \(\frac{55-r+1}{r} = \frac{2}{3}\). Solving this equation results in \(r = 28\).

Step 5: Identify the Terms

The values of \(r\) tell us that the equal coefficients occur in the 28th and 29th terms of the expansion.

Key Concepts

Binomial Coefficient

A binomial coefficient is a key component in the expansion of a binomial expression raised to a power. It's denoted by \(\binom{n}{k}\), which reads as "n choose k," and it signifies the number of ways to choose \(k\) items from a set of \(n\) items without regard to order. The formula to calculate the binomial coefficient is:

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

Here, \(!\) (factorial) means that you multiply a series of descending natural numbers.
For example, \(4!\) means \(4 \times 3 \times 2 \times 1\).

In our problem, these coefficients arise in the terms of the expansion derived from the binomial theorem. Understanding how they work is crucial because they determine the weight each term carries in the expansion of \((2+\frac{x}{3})^{55}\). This is foundational for finding those terms that have equal coefficients in the expansion.

Polynomial Expansion

Polynomial expansion is the process of breaking down a power expression, particularly a binomial, into a series of terms. Using the binomial theorem, this involves both the binomial coefficients and the powers of the individual terms in the binomial.

• The binomial theorem provides a shortcut to expand expressions like \((a + b)^n\).
• The general formula is: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \).

For the specific expression \(\left(2+\frac{x}{3}\right)^{55}\), every term in the expansion is composed of:

• A binomial coefficient \(\binom{55}{r-1}\).
• A power of the first term, \(2^{56-r}\).
• A power of the second term, \(\left(\frac{x}{3}\right)^{r-1}\).

This provides us with the structure needed to analyze and compute the power series, ultimately identifying terms with equal coefficients.

Equal Coefficients

Equal coefficients in polynomial expansions occur when the coefficients of consecutive terms in an expansion match. To find these, set up an equation where the coefficients of \(x^r\) and \(x^{r+1}\) are equal, which essentially looks like:

• \(\binom{55}{r-1} \cdot 2^{56-r} \cdot \frac{1}{3^{r-1}} = \binom{55}{r} \cdot 2^{55-r} \cdot \frac{1}{3^r}\)

Simplifying such an equation involves understanding both binomial coefficients and the rules of algebra. By solving the equation, you identify the value of \(r\) that makes the coefficients equal.

In this exercise, solving gives \(r = 28\), meaning the terms \(T_{28}\) and \(T_{29}\) in the expansion have equal coefficients. It's an interesting phenomenon and an important concept in algebra, as it shows symmetry and balance within polynomial expansions.