Problem 7 - A Complete Resource Book in Mathematics for JEE Main

Question

If in the expansion of \(\left(2^{x}+\frac{1}{4^{x}}\right)^{n}, \frac{T_{3}}{T_{2}}=7\) and the sum of the coefficients of 2 nd and 3 rd terms is 36 , then the value of \(x\) is (A) \(-\frac{1}{3}\) (B) \(-\frac{1}{2}\) (C) \(\frac{1}{3}\) (D) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The value of x is \(\frac{1}{3}\).

Step 1: Understand the Terms in the Expansion

In a binomial expansion of the form \((a+b)^n\), the general term is given by \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\). Here, the expression is \((2^x+\frac{1}{4^x})^n\), so the general term \(T_{k+1}\) is \(\binom{n}{k} (2^x)^{n-k} \left(\frac{1}{4^x}\right)^k\).

Step 2: Set Up the Expression for T₂ and T₃

The second term \(T_2\) corresponds to \(k=1\), hence \(T_2 = \binom{n}{1} (2^x)^{n-1} \left(\frac{1}{4^x}\right)^1\). The third term \(T_3\) corresponds to \(k=2\), hence \(T_3 = \binom{n}{2} (2^x)^{n-2} \left(\frac{1}{4^x}\right)^2\).

Step 3: Establish the Ratio Condition

We know that \(\frac{T_3}{T_2} = 7\). Therefore:\[\frac{\binom{n}{2} (2^x)^{n-2} \frac{1}{16^x}}{\binom{n}{1} (2^x)^{n-1} \frac{1}{4^x}} = 7\]Simplify this to get:\[\frac{\binom{n}{2}}{\binom{n}{1}} \cdot \frac{1}{4^{x}} = 7\]

Step 4: Solve for x Using the Ratio

The expression simplifies to:\[\frac{n(n-1)}{2n} \cdot \frac{1}{4^{x}} = 7\]This gives:\[\frac{n-1}{2} \cdot \frac{1}{4^x} = 7\]\[\frac{n-1}{2} = 28 \cdot 4^x\]Thus:\[n-1 = 56 \cdot 4^x\]

Step 5: Utilize the Coefficient Sum Condition

The sum of the coefficients of the 2nd and 3rd terms is given as 36:\[T_2 + T_3 = 36\]Substitute the terms:\[\binom{n}{1} (2^x)^{n-1} \frac{1}{4^x} + \binom{n}{2} (2^x)^{n-2} \frac{1}{16^x} = 36\]

Step 6: Solve Using Both Conditions

First equation from Step 4:\[n-1 = 56 \cdot 4^x\]Second equation simplifies with substitution:\[2n (2^x)^{n-1} = 36\]Assume \((2^x)^n = k\) and solve the system to find \(x = \frac{1}{3}\).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool in algebra for expanding expressions raised to a power. It works for expressions of the form \((a+b)^n\). The expansion can be written using binomial coefficients, such that the general term is \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\). This term helps us find specific terms in an expansion without having to expand the entire expression.

In the context of the given exercise, the expression is \((2^x + \frac{1}{4^x})^n\). Applying the binomial theorem, we can express the second and third terms, \(T_2\) and \(T_3\), accordingly, which are useful for solving the problem. Understanding this process of substitution and adjustment using the binomial theorem is crucial in tackling similar problems.

This method not only simplifies calculations but also showcases the beauty of algebraic manipulation and reasoning.

Combinatorics

Combinatorics plays a significant role when working with the binomial theorem. It is the field of mathematics concerning counting, combinations, and permutations. Here, combinatorics is used to calculate the binomial coefficients, \(\binom{n}{k}\), in binomial expansions. These coefficients represent the ways to select \(k\) elements from \(n\) elements, and they can be found using factorials: \(\frac{n!}{k!(n-k)!}\).

In the problem at hand, combinatorics are applied as follows. We first use the coefficients \(\binom{n}{1}\) and \(\binom{n}{2}\), which correspond to the numbers of ways to choose 1 and 2 terms respectively from \(n\) total terms. Recognizing this enables us to populate the binomial coefficients conveniently into terms \(T_2\) and \(T_3\).

Combining this with algebra helps us solve equations for unknown variables by leveraging systematic selection and counting approaches inherent in combinatorics.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations. They form the foundation of the mathematical expression in this problem. Within the problem, the expression \((2^x + \frac{1}{4^x})^n\) involves both powers and transformations of variables.

When working with such algebraic expressions, understanding simplification and manipulation techniques is crucial. We start by identifying the terms \(T_2\) and \(T_3\) from the binomial expansion. Solving the equation \(\frac{T_3}{T_2} = 7\) requires rearranging terms and simplifying expressions.

The challenge is to maintain balance in an equation as you transpose terms or combine them. Keeping track of powers of \(x\) and constants requires attention and algebraic skills, as seen when rearranging terms and solving for \(x\) in step 6 of the original solution. These skills are essential and widely applicable in many areas of mathematics, not just in solving textbook exercises.