Problem 123 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(3^{1 / 3} \cdot 9^{1 / 9} \cdot 27^{1 / 27} \cdot 81^{1 / 81} \ldots\) upto \(\infty=\) (A) \(\sqrt{27}\) (B) \(\sqrt[3]{27}\) (C) \(\sqrt[4]{27}\) (D) None of thesePassage 3 We know that arithmetic mean of the positive numbers lie between them. Suppose, \(S_{1}>S_{2}>0\) and \(S_{n+1}=\frac{1}{2}\left(S_{n}+S_{n-1}\right)\) We can easily conclude that \(S_{3}\) lies between \(S_{2}\) and \(S_{1}\) and we can write \(S_{2}

Step-by-Step Solution

Answer

The product converges to option C, \(\sqrt[4]{27}\).

Step 1: Understanding Exponential Forms

First, let's rewrite each term using exponentials: \[3^{1/3} = 3^{1/3}\]\[9^{1/9} = (3^2)^{1/9} = 3^{2/9}\]\[27^{1/27} = (3^3)^{1/27} = 3^{3/27} = 3^{1/9}\]\[81^{1/81} = (3^4)^{1/81} = 3^{4/81}\]Thus, each term in the product sequence is expressed in terms of base 3 with fractional exponents.

Step 2: Combining the Exponentials

The entire expression that we're evaluating is:\[3^{1/3} \cdot 3^{2/9} \cdot 3^{1/9} \cdot 3^{4/81} \cdots\]We can combine the exponential terms by summing their exponents:\[3^{1/3 + 2/9 + 1/9 + 4/81 + \cdots}\]

Step 3: Observing the Infinite Sum

Notice that the exponent of 3 in the product forms an infinite series. Each exponent can be viewed as:\[\frac{1}{1 \cdot 3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \cdots\]This is a known series where the nth term is \(\frac{n}{3^n}\). Such series is known to converge.

Step 4: Calculating the Sum of the Series

Using calculus, the sum of the series can be found as:\[\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{x}{(1-x)^2} \quad \text{where} \quad x = \frac{1}{3}\]Plugging in for \(x\):\[\frac{\frac{1}{3}}{(1-\frac{1}{3})^2} = \frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{9}{12} = \frac{3}{4}\]Thus, the series converges to \(\frac{3}{4}\).

Step 5: Evaluating the Total Expression

Now that we know the sum of the exponents, the original infinite product simplifies to:\[3^{\sum_{n=1}^{\infty} \frac{n}{3^n}} = 3^{3/4}\]This can be simplified to \(\sqrt[4]{27}\), which corresponds to option C.

Key Concepts

Exponential Forms

Understanding exponential forms is essential when dealing with expressions that involve powers and roots. In mathematics, exponential expressions, such as \(3^{1/3}\) or \(9^{1/9}\), take a base number and raise it to a power or an exponent. This power can be an integer, a fraction, or even a negative number. For instance, in our problem, we are presented with a sequence of terms like \(3^{1/3}\), \((3^2)^{1/9}\), \((3^3)^{1/27}\), and so forth. In each case, these can be rewritten using the properties of exponents: \((a^m)^n = a^{m imes n}\). This allows us to simplify expressions like \(9^{1/9} = (3^2)^{1/9} = 3^{2/9}\) back to the base of 3. So, through understanding exponential forms, we can recognize patterns and simplify complex mathematical expressions into simpler, unified forms, which in this case all share the base 3.

Infinite Series

An infinite series is a sum of an infinite sequence of terms. It is written as \(a_1 + a_2 + a_3 + \cdots\). In our problem, the series appears in the form of the exponents of the base 3 terms. These exponents are each part of an infinite sum: \(\frac{1}{1 \cdot 3} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots\). Such expressions can be complex because you are effectively adding together an infinite number of numbers. Often with infinite series, we look to find whether the sum converges to a specific value or not. The series here is comprised of terms in the form \(\frac{n}{3^n}\), indicating it follows a specific pattern: each term gets smaller as \(n\) increases due to the denominator exponentially growing. Hence, learning about infinite series helps us recognize these patterns and interpret them in sums that can stretch on indefinitely.

Convergent Series

A convergent series is an infinite series whose terms approach a certain number as more and more terms are added. Simply put, it's when adding up an infinite sequence of numbers leads to a specific limit, rather than continuing on to infinity. In the example provided, the series \(\sum_{n=1}^{\infty} \frac{n}{3^n}\) is noted to be convergent. This means that as we continue to add more terms, the sum approaches a particular value, which through calculus, we have calculated to be \(\frac{3}{4}\). This is important because it confirms that the seemingly infinite problem can actually yield a finite and meaningful result. Convergence is a foundational concept in understanding how infinite products might "behave" as they stretch infinitely. Without such principles, it would be impossible to accurately determine outcomes in infinite series practically.