Problem 4 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots\) to \(n\) terms \(=\) (A) \(\frac{1}{5 n-1}\) (B) \(\frac{1}{3 n-1}\) (C) \(\frac{n}{3 n+1}\) (D) \(\frac{n}{5 n-1}\)

Step-by-Step Solution

Answer

The answer is (C) \(\frac{n}{3n+1}\).

Step 1: Identify the General Term

First, let's identify the general term of the series. Notice that the series is expressed as a sequence of fractions: \( \frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \ldots \). Here, each denominator is the product of terms that increase by 3. Thus, the \(n\)-th term can be written as \( \frac{1}{(3n-2)(3n+1)} \).

Step 2: Apply Partial Fraction Decomposition

To simplify the general term \( \frac{1}{(3n-2)(3n+1)} \), use partial fraction decomposition. Assume \( \frac{1}{(3n-2)(3n+1)} = \frac{A}{3n-2} + \frac{B}{3n+1} \). Solving for \(A\) and \(B\), we get:\[1 = A(3n+1) + B(3n-2).\] Setting up the equations: \[ A + B = 0 \quad \text{and} \quad 3A - 2B = 1 \]Solving gives \( A = \frac{2}{3} \) and \( B = -\frac{2}{3} \). Thus, the term becomes:\[\frac{1}{3} \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right).\]

Step 3: Recognize the Telescoping Nature

The series with general term \( \frac{1}{3} \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right) \) is telescoping. When you expand the series for various terms, each fraction cancels out the subsequent part of another fraction, leaving only the first and last parts.

Step 4: Write the Expansion

Write the expansion of the series up to \(n\) terms:\[ \frac{1}{3} \left( \frac{1}{1} - \frac{1}{4} \right) + \frac{1}{3} \left( \frac{1}{4} - \frac{1}{7} \right) + \ldots + \frac{1}{3} \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right).\] Notice how each term cancels except for a few initial and final fractions.

Step 5: Simplify the Result

The telescoping nature leaves you with only the terms:\[\frac{1}{3} \left( \frac{1}{1} - \frac{1}{3n+1} \right) = \frac{1}{3} \left( 1 - \frac{1}{3n+1} \right).\]Simplify to find the sum:\[\frac{1}{3} \left( \frac{3n+1-1}{3n+1} \right) = \frac{n}{3n+1}.\]

Step 6: Choose the Correct Answer

From the simplified form \( \frac{n}{3n+1} \), identify it as option (C) in the provided choices.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in algebra to break down complex rational expressions into simpler, more manageable parts. This method is particularly useful when dealing with polynomials in the denominator. In many cases, it's difficult to integrate or simplify expressions like \( \frac{1}{(3n-2)(3n+1)} \). Using partial fraction decomposition allows us to express it in the form \( \frac{A}{3n-2} + \frac{B}{3n+1} \) for some constants \( A \) and \( B \).To find these constants, we first equate the original expression to the sum of the simpler expressions. Here, that gives us:

• \( 1 = A(3n+1) + B(3n-2) \)

By expanding and rearranging this equation, we determine the values of \( A \) and \( B \). We found \( A = \frac{2}{3} \) and \( B = -\frac{2}{3} \). The original fraction thus simplifies to \( \frac{1}{3} ( \frac{1}{3n-2} - \frac{1}{3n+1} ) \). This decomposition simplifies complex algebraic manipulations, particularly in evaluating series.

Telescoping Series

A telescoping series is one where the terms of the series cancel out in such a way that it becomes easy to find the sum. Each term essentially "telescopes" into the next, meaning many terms will cancel out each other, leaving a much simpler expression.In our specific example, the series can be written with the general term: \( \frac{1}{3} \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right) \). As you expand this series from the first term to the \( n \)-th term:

• First term: \( \frac{1}{3} ( \frac{1}{1} - \frac{1}{4} ) \)
• Second term: \( \frac{1}{3} ( \frac{1}{4} - \frac{1}{7} ) \)
• And so on, up to \( n \)-th term: \( \frac{1}{3} ( \frac{1}{3n-2} - \frac{1}{3n+1} ) \)

Each intermediate fraction in these terms cancels with part of the preceding or the following one. Eventually, almost all terms cancel, leaving only the first positive part of the first term and the negative part of the last term. This cancellation leads to a simple formula for summing the series.

Sequence and Series

The terms 'sequence' and 'series' are fundamental concepts in mathematics. A sequence is an ordered list of numbers, each of which is called a term. A series, on the other hand, is the sum of terms of a sequence. In our example, the sequence is expressed as \( \frac{1}{1 \times 4}, \frac{1}{4 \times 7}, \frac{1}{7 \times 10}, \ldots \). The series is the sum of this sequence up to \( n \) terms. Sequences can be finite or infinite, and series can converge to a limit or diverge.Understanding sequences and series is crucial in various fields like calculus and analysis. They form the groundwork for defining concepts such as convergence, limits, and continuity. Furthermore, the analysis of sequences and series allows for solving complex real-world problems in engineering, physics, and finance.In our specific case, recognizing the structure of our series allowed us to use partial fraction decomposition and recognize telescoping characteristics. As a result, we effectively simplified and calculated the sum for the given sequence up to \( n \) terms.