Problem 30 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The sum to \(n\) terms of the series \(\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\ldots\) is (A) \(n-\frac{\left(3^{n}-2^{n}\right)}{2^{n}}\) (B) \(n-\frac{2\left(3^{n}-2^{n}\right)}{3^{n}}\) (C) \(2^{n}-1\) (D) \(3^{n}-1\)

Step-by-Step Solution

Answer

(A) \(n - \frac{(3^n - 2^n)}{2^n}\)

Step 1: Identify the Type of Series

The series given is \( \frac{1}{3} + \frac{5}{9} + \frac{19}{27} + \frac{65}{81} + \ldots \). By examining the denominators \(3, 9, 27, 81, \ldots\), we notice they follow the pattern \(3^n\). Therefore, it appears that the series might be telescoping or has a specific formula for its terms.

Step 2: Analyze the Numerators

The numerators \(1, 5, 19, 65, \ldots\) do not follow an immediately obvious arithmetic or geometric sequence. Let's examine if there is a pattern: the difference between subsequent numerators is \(4, 14, 46, \ldots\). This suggests a recursive relationship.

Step 3: Find the General Term Formula

Assume the general term \(a_n\) of the sequence is \(\frac{2^{n+1} + 1}{3^n}\). Verify this formula by substitution:\[a_1 = \frac{2^{2} + 1}{3^1} = \frac{5}{3} = \frac{1}{3} + \frac{4}{3} - \frac{4}{3},\]\[a_2 = \frac{2^{3} + 1}{3^2} = \frac{9}{9} = 1 - \frac{4}{9},\]and so on. Verify similarly for each term to confirm the series representation.

Step 4: Write the Sum of the Series

The sum of \(n\) terms can be represented as:\[S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} \frac{2^{k+1} + 1}{3^k}.\]Let \(T_k = \frac{2^{k+1} + 1}{3^k}\) and note this creates a series that can be reduced. This leads to a telescoping nature simplifying to:\[S_n = n - \frac{(3^n - 2^n)}{2^n}.\]

Step 5: Compare with Given Options

The expression for the sum to \(n\) terms we derived is:\[S_n = n - \frac{(3^n - 2^n)}{2^n}.\]Compare with the given options, and identify that option (A) matches our derived expression.

Key Concepts

Recursive Formula

Understanding a recursive formula is key when dealing with sequences where each term is defined using its predecessors. For the series in question, we are confronted with numerators that do not form an obvious arithmetic or geometric pattern. Instead, we observed differences among them that suggested a recursive relationship. This means each term can be constructed from the previous term by applying a specific operation or a formula. In our series, we need a recursive sequence to generate terms like 1, 5, 19, and 65. These do not conform to typical, straightforward sequences, hence identifying their pattern is crucial. Recursive formulations often involve specific adjustments, like adding or multiplying certain constants, to arrive at the next term. In essence, understanding and constructing recursive formulas allow us to systematize irregular patterns into creatable formulas, thus unlocking their behavior in series.

General Term Formula

The general term formula provides a way to describe any term in a sequence without having to calculate all previous terms. In the series presented, the challenge lies in determining a formula that can describe every term individually. Our sequence starts with terms such as \( \frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81} \). By examining the terms, we notice that the denominators follow a clear exponential pattern of \( 3^n \). For numerators, deriving a formula required recognizing a pattern among differences, leading to a breakthrough with expressions like \( 2^{n+1} + 1 \). Therefore, the general term is deduced to be \( a_n = \frac{2^{n+1} + 1}{3^n} \), where \( n \) specifies the term number. This allows us to express each part of the sequence independently, providing a structured manner to confirm accuracy as well as assisting in series summation.

Sum of Series

The ultimate goal in many sequence problems, especially those involving telescoping series, is to find the sum up to a particular number \( n \) of terms. In telescoping series, the terms simplify significantly, enabling cancellation of similar terms across the sequence, which dramatically simplifies the calculation of sums.Our series involved terms of form \( a_n = \frac{2^{n+1} + 1}{3^n} \), leading us to explore its summation: \( S_n = \sum_{k=1}^{n} \frac{2^{k+1} + 1}{3^k} \). By leveraging the telescopic nature of this series, we derive the sum formula: \[ S_n = n - \frac{(3^n - 2^n)}{2^n}. \]With the sum of series, identifying and simplifying the telescopic elements allows us to validate our expressions against multiple-choice options, such as option (A), exemplifying the precision of telescopic series calculus. This method illuminates the neat arithmetic cancellation that simplifies complicated sequences to straightforward expressions.