Problem 21 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the coefficients of \(\mathrm{x}^{-2}\) and \(\mathrm{x}^{-4}\) in the expansion of \(\left(x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{1}{3}}}\right)^{18},(x>0)\), are \(m\) and \(n\) respectively, then \(\frac{m}{n}\) is equal to : (a) 27 (b) 182 (c) \(\frac{5}{4}\) (d) \(\frac{4}{5}\)

Step-by-Step Solution

Answer

The ratio \(\frac{m}{n}\) is 27, option (a).

Step 1: Apply the Binomial Theorem

Consider the binomial expansion \( \left( x^{\frac{1}{3}} + \frac{1}{2x^{\frac{1}{3}}} \right)^{18} \). According to the binomial theorem, the general term \( T_k \) in the expansion is given by:\[ T_k = \binom{18}{k} \left(x^{\frac{1}{3}}\right)^{18-k} \left(\frac{1}{2x^{\frac{1}{3}}}\right)^k. \]

Step 2: Simplify the General Term

Simplifying the expression for the general term, we have:\[ T_k = \binom{18}{k} (x^{\frac{1}{3}(18-k)}) \left(\frac{1}{2}\right)^k (x^{-\frac{k}{3}}) = \binom{18}{k} \frac{1}{2^k} x^{\frac{18-k-k}{3}}. \]This simplifies to:\[ T_k = \binom{18}{k} \frac{1}{2^k} x^{\frac{18-2k}{3}}. \]

Step 3: Find the Coefficient of \(x^{-2}\)

We need \((\frac{18-2k}{3}) = -2\). Solving for \(k\):\[ \frac{18-2k}{3} = -2 \rightarrow 18 - 2k = -6 \rightarrow 2k = 24 \rightarrow k = 12. \]Substituting \(k = 12\) into the general term, the coefficient of \(x^{-2}\) is:\[ \binom{18}{12} \frac{1}{2^{12}}. \]Term this coefficient as \(m\).

Step 4: Find the Coefficient of \(x^{-4}\)

We need \((\frac{18-2k}{3}) = -4\). Solving for \(k\):\[ \frac{18-2k}{3} = -4 \rightarrow 18 - 2k = -12 \rightarrow 2k = 30 \rightarrow k = 15. \]Substituting \(k = 15\) into the general term, the coefficient of \(x^{-4}\) is:\[ \binom{18}{15} \frac{1}{2^{15}}. \]Term this coefficient as \(n\).

Step 5: Calculate \(\frac{m}{n}\)

We compute the ratio of the coefficients:\[ \frac{m}{n} = \frac{\binom{18}{12} \frac{1}{2^{12}}}{\binom{18}{15} \frac{1}{2^{15}}} = \frac{\binom{18}{12}}{\binom{18}{15}} \times 2^3. \]Using the relation \(\binom{n}{k} = \binom{n}{n-k}\), we know \(\binom{18}{12} = \binom{18}{6}\) and \(\binom{18}{15} = \binom{18}{3}\), thus the ratio becomes:\[ \frac{\binom{18}{6}}{\binom{18}{3}} \times 8. \]Calculating,\[ \binom{18}{6} = 18564\] and \[ \binom{18}{3} = 816\], thus:\[ \frac{18564}{816} \times 8 = 27. \]

Step 6: Select the Correct Option

Therefore, the correct answer is \(27\), which corresponds to option (a).

Key Concepts

General Term

In the context of the binomial expansion, the general term is a vital concept. It allows us to express each term in the expanded form of a binomial raised to a power. To determine a specific term within the expansion, we use the general term formula, which incorporates several elements:

• Binomial coefficient: This signifies the number of ways to choose elements in a set, represented as \( \binom{n}{k} \).
• Variable powers: Each term includes specific powers of variables as dictated by the problem.

For example, in the exercise \( \left( x^{\frac{1}{3}} + \frac{1}{2x^{\frac{1}{3}}} \right)^{18} \), the general term \( T_k \) is expressed as:\[ T_k = \binom{18}{k} \left(x^{\frac{1}{3}}\right)^{18-k} \left(\frac{1}{2x^{\frac{1}{3}}}\right)^k \]This formula enables us to find specific terms by substituting different values for \( k \). This concept becomes especially critical when searching for terms with specific powers, such as \( x^{-2} \) or \( x^{-4} \). By adjusting \( k \), one finds the desired specific terms in the binomial expansion.

Binomial Coefficients

Binomial coefficients appear in the general term of the binomial expansion and serve a key role. They are all about counting combinations. Essentially, the binomial coefficient \( \binom{n}{k} \) tells us the number of ways to select \( k \) items from \( n \) items, without regard for the order of selection. These coefficients can be computed using the formula:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]Where \( n! \) is the factorial of \( n \), and it represents the product of all positive integers up to \( n \).
In our exercise, computing the coefficients is necessary for finding the values of \( m \) and \( n \). For instance, we calculated:

• \( \binom{18}{12} \) to find the coefficient of \( x^{-2} \)
• \( \binom{18}{15} \) for the coefficient of \( x^{-4} \)

These coefficients eventually help in evaluating the ratio \( \frac{m}{n} \), which demands precise calculation of these binomial coefficients. Understanding how to effectively compute and use these coefficients is essential to decipher any binomial problem.

Exponent Manipulation

Exponent manipulation is a critical skill, especially when dealing with binomial expansions. This skill involves working with powers of a variable in a way that simplifies expressions and equations. In the given exercise, exponent manipulation is used extensively to solve for the specific terms:

• We start with terms such as \( x^{\frac{1}{3}(18-k)} \) and \( x^{-\frac{k}{3}} \)
• Next, these are combined into a simplified power of \( x \), resulting in \( x^{\frac{18-2k}{3}} \)

To find the coefficients of specific power terms such as \( x^{-2} \) and \( x^{-4} \), it's crucial to set up the exponent equation correctly:\[ \frac{18-2k}{3} = -2 \hspace{1cm} \text{or} \hspace{1cm} \frac{18-2k}{3} = -4 \]Solving these equations for \( k \) allows us to determine which term of the binomial expansion produces the desired power of \( x \). This manipulation transforms even complex terms into manageable parts, making the entire process straightforward and solvable.