Problem 138 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \ldots \infty\) is: \(\quad[2002]\) (A) 1 (B) 2 (C) \(3 / 2\) (D) 4

Step-by-Step Solution

Answer

The value is 2.

Step 1: Understand the Problem

We are required to find the value of the infinite product: \(2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \ldots \infty\). This sequence consists of powers of even numbers with decreasing exponents.

Step 2: Express Terms Using Base 2

Note that each number in the sequence can be expressed as a power of 2. Specifically: \(4 = 2^2\), \(8 = 2^3\), and so forth. Therefore, the sequence can be rewritten as: \(2^{1/4} \cdot (2^2)^{1/8} \cdot (2^3)^{1/16} \cdots\)

Step 3: Simplify Exponents

Distribute the exponents through each factor, giving us \(2^{1/4} \cdot 2^{2/8} \cdot 2^{3/16} \cdots\). Further simplifying, \(2^{2/8} = 2^{1/4}\) and \(2^{3/16}\) can be calculated similarly.

Step 4: Recognize the Pattern

Notice that each exponent is following a pattern: \(\frac{1}{4}, \frac{2}{8}, \frac{3}{16}, \ldots\). In general, the \(n\)-th term can be represented as \(\frac{n}{2^{n+1}}\). This is a geometric series in summation form.

Step 5: Sum the Infinite Series

The exponents to find the total exponent of the power of 2 would be \(\sum_{n=1}^\infty \frac{n}{2^{n+1}}\). Recognize that this series can be rearranged by factoring out \(1/2\) and becomes \(\frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^n}\), a known series with sum \(2\). Therefore, the net exponent becomes \(\frac{1}{2} \times 2 = 1\).

Step 6: Calculate the Value of the Expression

As the exponent of 2 is 1, the value of the expression is \(2^1 = 2\).

Key Concepts

Geometric Series

A geometric series is a series with a constant ratio between successive terms. It is an important concept in mathematics that is often used to describe and solve problems involving repeated multiplication by a constant factor. In general, a geometric series can be written as:

• a, ar, ar^2, ar^3, ..., ar^n, ...
• where \(a\) is the first term and \(r\) is the common ratio.

This concept is powerful because it allows us to easily calculate the sum of terms, even when the series is infinite. If \(|r| < 1\), the sum \(S\) of an infinite geometric series can be found using the formula: \[ S = \frac{a}{1 - r} \]This formula helps simplify complex problems, like the one in our exercise, by converting an infinite product into something more manageable. Here, it is the structure of the exponents that form a geometric series sequence.

Exponents

Exponents are used to express repeated multiplication of a number by itself. For example, \(2^3 = 2 \times 2 \times 2 = 8\). In the context of this exercise, every term in the sequence is a power with a decreasing exponent. To work more easily with exponents:

• Remember the rules of exponents such as \(a^m \times a^n = a^{m+n}\) and \((a^m)^n = a^{m \times n}\).
• Be comfortable converting different bases into the same base where possible, making calculations simpler.

In solving our infinite product, understanding how to simplify expressions by adjusting and combining exponents is key. This systematizes the sequence and helps us derive the meaningful pattern that guides the solution.

Series Summation

Series summation involves adding up a sequence of numbers. In mathematics, when dealing with series, especially infinite ones, it is important to determine if the series converges to a finite value.

• A series converges if the sum of its terms approaches a fixed number as you add more and more terms.
• Many known series, like the geometric series, have specific formulas for easy summation.

For the given problem, we used series summation to find the combined value of all exponents—by summing \(\frac{n}{2^{n+1}}\), a well-known series that converges to 2.By recognizing and summing this series, it gives us the total effective exponent of 2 required to compute the overall value of the infinite product. This clever technique changes a seemingly unmanageable infinite sequence into a solvable result.