Problem 55 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The product \(2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \ldots\) to \(\infty\) is equal to: [Jan. 9, 2020 (I)] (a) \(2^{\frac{1}{2}}\) (b) \(2^{\frac{1}{4}}\) (c) 1 (d) 2

Step-by-Step Solution

Answer

(a) \(2^{\frac{1}{2}}\)

Step 1: Express each term as a power of 2

Each term in the product is a power of 2:
\(2^{\frac{1}{4}}, \quad 4^{\frac{1}{16}} = 2^{\frac{2}{16}} = 2^{\frac{1}{8}}, \quad 8^{\frac{1}{48}} = 2^{\frac{3}{48}} = 2^{\frac{1}{16}}, \quad 16^{\frac{1}{128}} = 2^{\frac{4}{128}} = 2^{\frac{1}{32}}, \ldots\)
The general \(n\)-th term is \((2^n)^{\frac{1}{n \cdot 4^n}} \cdot \ldots\). Let us identify the pattern: the exponents of 2 are \(\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \ldots\)

Step 2: Identify the sum of exponents

The product equals \(2^S\) where \(S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots\)
This is a geometric series with first term \(a = \frac{1}{4}\) and common ratio \(r = \frac{1}{2}\).

Step 3: Compute the sum of the geometric series

The sum of an infinite geometric series is \(S = \frac{a}{1-r} = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}\).

Step 4: Find the product

The infinite product equals \(2^S = 2^{\frac{1}{2}} = \sqrt{2}\).
The answer is (a) \(2^{\frac{1}{2}}\).

Key Concepts

Series

A series is a sum of terms that are typically part of a sequence. In mathematics, when we talk about series, we usually mean infinite series, where infinite terms are added together. In the given problem, however, we're dealing with an infinite product, which is similar but involves multiplication instead of addition.
Understanding series is crucial since they are foundational in calculus and analysis. Series can be defined as convergent or divergent. Convergent series result in a finite value as more terms are added, while divergent series continue to grow indefinitely.
An important concept related to series is the idea of a limit. The limit of a sequence is a value the sequence approaches as the number of terms increases. Convergence in series is often determined by the behavior of this limit. For example, an infinite geometric series with a common ratio between (-1 and 1) converges to a finite value, given by the formula \( S = \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio.

Infinite Product

An infinite product extends the idea of multiplication over an infinite number of factors. In the context of the exercise, we deal with powers of 2. When you encounter an infinite product like this, it's helpful to simplify each part using a common base, as done in the solution.
The infinite product paradigm focuses on whether the product of an infinite number of terms converges or diverges. It converges if the product results in a number that stabilizes as you multiply more terms.
In this specific problem, each term is expressed as \(2^{\frac{n}{a}}\), gradually forming exponents that tell us how multiplication progresses. The challenge is to understand that such products can also reach a limit, where, in this instance, the product results in \(2^{\frac{1}{2}}\).
To identify convergent behavior, one can investigate individual exponents' fractional components. As they become smaller, the resulting effect might stabilize, indicating convergence.

Base Conversion

Base conversion is an important mathematical skill, used here to express the entire product in terms of powers of 2. The aim is to simplify the expression by converting numbers into a consistent base.
When faced with exponents involving different bases, one strategy is to convert all numbers into a single base. In this problem, all bases are turned into powers of 2. For instance, \(4 = 2^2\), \(8 = 2^3\), and \(16 = 2^4\), transforming these into terms like \( 2^{\frac{1}{8}} \) and \( 2^{\frac{1}{16}} \).
Base conversion helps in comparing, adding, or multiplying terms with different original bases by placing them on a common ground, making complicated expressions manageable. It is especially useful when dealing with sequences or products where terms need to interact algebraically, ensuring calculations can be carried out more efficiently.