Problem 118 - A Complete Resource Book in Mathematics for JEE Main

Question

Sum to \(n\) terms of the series \(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots .\) is (A) \(\frac{n}{2 n+1}\) (B) \(\frac{n}{2 n-1}\) (C) \(\frac{n-1}{2 n+1}\) (D) None of these

Step-by-Step Solution

Answer

The sum to \( n \) terms is \( \frac{n}{2n+1} \), option (A).

Step 1: Identify the pattern

The given series is \( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots \). We observe that each term has the form \( \frac{1}{(2k-1)(2k+1)} \) for \( k = 1, 2, 3, \ldots \). This pattern is a series of fractions where the numerators are consistently 1 and the denominators are consecutive odd integers multiplied together.

Step 2: Apply partial fraction decomposition

We can express \( \frac{1}{(2k-1)(2k+1)} \) as a difference of two fractions: \( \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \). This identity allows us to rewrite each term so that when summed, many terms will cancel with adjacent terms.

Step 3: Simplify the series

Substitute the decomposition from Step 2 into the series:\[\sum_{k=1}^{n} \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right)\]This is a telescoping series. Most terms will cancel out, leaving:\[\sum_{k=1}^{n} \frac{1}{2} (1 - \frac{1}{2n+1}) \] Upon simplification, this becomes:\( \frac{1}{2} \left( 1 - \frac{1}{2n+1} \right) = \frac{n}{2n+1} \).

Step 4: Identify the correct answer from options

The final simplified form of the sum is \( \frac{n}{2n+1} \). Looking at the provided options, the correct choice is (A) \( \frac{n}{2n+1} \).

Key Concepts

Series Summation

Series summation involves adding terms of a sequence to find their total. In our exercise, we're dealing with a special type of series called a telescoping series. This series has terms that, when summed, simplify due to cancellation among the terms. A telescoping series typically looks like this: \[ \sum_{k=1}^{n} a_k \]Where each term \( a_k \) can be expressed as the difference of two consecutive fractions. During the summation process, most intermediate terms cancel out, making the calculation much more manageable.
For instance, in the given exercise, each term can be rewritten using partial fraction decomposition, leading it to a form where consecutive terms cancel each other, except for the first and the last part of the series when summed. This characteristic simplifies solving it without explicit computation of each individual term.Understanding the process of summation in such series not only simplifies solving exercises but also helps in recognizing patterns, as many complex series problems can be broken down using similar methods.

Partial Fraction Decomposition

Partial fraction decomposition is a powerful technique used in calculus to simplify complex fractional expressions. The goal is to express a fraction as a sum or difference of simpler fractions, which are easier to handle when summing up the terms of a series.For the series exercise given, each term before summing can be expressed as:\[ \frac{1}{(2k-1)(2k+1)} = \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \]This identity transforms the original terms into simpler components that facilitate cancellation. Partial fraction decomposition often takes advantage of differences like this to transform a product of polynomials in the denominator into separated terms, making telescoping possible.To master this technique:

• Identify the structure of the denominator as a product of linear factors, if possible.
• Express the fraction using constants (like our \(\frac{1}{2}\)) that enable cancellation when summed.
• Recognize the resulting pattern to simplify telescoping later on.

With practice, partial fraction decomposition reveals an elegant way to solve problems involving series and integrals.

Mathematical Patterns

Recognizing mathematical patterns is a vital skill when solving complex mathematical problems, particularly those involving series. Identifying patterns enables you to choose appropriate mathematical techniques, such as partial fraction decomposition or series summation, to simplify and solve the problem efficiently.In the context of the exercise, spotting the odd number pattern in the denominators, \((2k-1)(2k+1)\), offers a clue. These are sequential odd numbers, which helps foresee that the series might telescope.The pattern continues in the form of a difference of fractions after decomposition, visible in the step:\[ \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \]This reinforces that terms cancel out systematically as they sum over \(n\) terms, essentially leaving a trivial sum of the first and an adjusted last part of the sequence.
Knowing how to identify such patterns simplifies problem-solving and reduces complexity. When preparing for examinations or homework, break down the expression to identify familiar patterns and apply the correct strategies accordingly.