Product Of Sequence: (1/3)*(2/4)*...*(98/100) Solution

Let's dive into how to evaluate the product of the sequence: ( \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \dots \cdot \frac{97}{99} \cdot \frac{98}{100} ). This problem involves multiplying a series of fractions, and at first glance, it might seem daunting. However, by understanding the pattern and applying some clever mathematical techniques, we can simplify it and arrive at the solution. In this article, we'll break down the sequence, identify the pattern, and use a step-by-step approach to find the product. So, buckle up and let's get started!

Understanding the Sequence

Before we jump into calculations, it's crucial to understand the nature of the sequence we're dealing with. The sequence is a product of fractions, where each fraction has the form ${ \frac{n}{n+2} }$, and ${ n }$ ranges from 1 to 98. Writing out the first few and last few terms can help visualize the sequence:

${ \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{4}{6} \cdot \dots \cdot \frac{95}{97} \cdot \frac{96}{98} \cdot \frac{97}{99} \cdot \frac{98}{100} }$

Notice how each numerator is two less than its denominator. This pattern suggests there might be some cancellation of terms when we write out the entire product. Spotting these patterns is key to simplifying complex mathematical expressions.

Identifying the Pattern and Simplification Strategy

The key to solving this problem lies in recognizing the potential for cancellation. When we multiply these fractions together, some terms in the numerators will cancel with terms in the denominators. To better illustrate this, let's write out the product in an expanded form:

${ \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \dots \cdot 95 \cdot 96 \cdot 97 \cdot 98}{3 \cdot 4 \cdot 5 \cdot 6 \cdot \dots \cdot 97 \cdot 98 \cdot 99 \cdot 100} }$

Now, we can see clearly which terms cancel out. The numerator contains the product of integers from 1 to 98, and the denominator contains the product of integers from 3 to 100. Most of the terms will cancel each other out, leaving only a few terms at the beginning of the numerator and at the end of the denominator.

Performing the Cancellation

Upon closer inspection, we observe that the terms from 3 to 98 appear in both the numerator and the denominator. Thus, we can cancel them out. After cancellation, we're left with:

${ \frac{1 \cdot 2}{99 \cdot 100} }$

This simplifies to:

${ \frac{2}{9900} }$

Final Calculation and Result

Now, we just need to simplify the fraction ${ \frac{2}{9900} }$. We can divide both the numerator and the denominator by 2:

${ \frac{2 \div 2}{9900 \div 2} = \frac{1}{4950} }$

So, the final result of the product of the sequence is:

${ \frac{1}{4950} }$

Therefore, ${ \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \dots \cdot \frac{97}{99} \cdot \frac{98}{100} = \frac{1}{4950} }$.

Alternative Approach: Gamma Function (Advanced)

For those familiar with advanced mathematical concepts, an alternative approach involves the Gamma function. The Gamma function is a generalization of the factorial function to complex numbers. While it's a more complex method, it offers a different perspective on solving this type of problem.

The product can be written in terms of factorials as follows:

${ \prod_{n=1}^{98} \frac{n}{n+2} = \frac{\prod_{n=1}^{98} n}{\prod_{n=1}^{98} (n+2)} = \frac{98!}{\frac{100!}{2!}} = \frac{98! \cdot 2}{100!} = \frac{2}{99 \cdot 100} = \frac{2}{9900} = \frac{1}{4950} }$

This approach confirms our earlier result.

Common Mistakes to Avoid

When evaluating such sequences, it's easy to make a few common mistakes:

1. Incorrect Cancellation: Ensure you correctly identify which terms cancel out. A small error in identifying the range of cancellation can lead to a completely wrong answer.
2. Arithmetic Errors: Double-check your arithmetic, especially when simplifying fractions. Simple calculation mistakes can be easily avoided with careful attention.
3. Forgetting the Pattern: Always remember the underlying pattern of the sequence. Understanding the pattern is critical for setting up the problem correctly.

Conclusion

Evaluating the product of the sequence ${ \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \dots \cdot \frac{97}{99} \cdot \frac{98}{100} }$ involves recognizing the pattern, simplifying the expression through cancellation, and performing basic arithmetic. By carefully following these steps, we found that the product equals ${ \frac{1}{4950} }$. Whether you're a student tackling a math problem or a math enthusiast, understanding these techniques can significantly enhance your problem-solving skills. Keep practicing, and you'll become more comfortable with these types of mathematical challenges! Hey guys, hope you found this helpful!

Additional Tips for Sequence and Series Problems

To further enhance your skills in solving sequence and series problems, consider the following tips:

• Write Out Terms: Whenever you encounter a sequence or series, writing out the first few terms can help you visualize the pattern and understand the behavior of the sequence.
• Look for Differences or Ratios: Check if the sequence has a constant difference (arithmetic sequence) or a constant ratio (geometric sequence) between consecutive terms. This can simplify the analysis.
• Telescoping Series: Be on the lookout for telescoping series, where intermediate terms cancel out, leaving only the first and last few terms.
• Partial Fractions: If you're dealing with a series of rational functions, consider using partial fraction decomposition to simplify the terms.
• Induction: Mathematical induction can be a powerful tool for proving properties of sequences and series, especially when you have a recursive definition.
• Practice Regularly: The more you practice, the better you'll become at recognizing patterns and applying appropriate techniques. Solve a variety of problems to broaden your skills.

By incorporating these tips into your problem-solving approach, you'll be well-equipped to tackle a wide range of sequence and series problems. Remember, patience and persistence are key!

Real-World Applications of Sequences and Series

Sequences and series aren't just abstract mathematical concepts; they have numerous applications in the real world. Here are a few examples:

1. Finance: Compound interest, annuities, and loan amortization all involve sequences and series. Understanding these concepts is crucial for making informed financial decisions.
2. Physics: Many physical phenomena, such as oscillations, waves, and radioactive decay, can be modeled using sequences and series. For example, Fourier series are used to analyze complex waveforms.
3. Computer Science: Sequences and series are used in algorithms, data structures, and numerical analysis. For example, the efficiency of certain algorithms can be analyzed using recurrence relations, which are a type of sequence.
4. Engineering: Engineers use sequences and series in various applications, such as signal processing, control systems, and structural analysis. For example, Taylor series are used to approximate functions and solve differential equations.
5. Biology: Population growth, genetic inheritance, and enzyme kinetics can be modeled using sequences and series. For example, the Fibonacci sequence appears in the arrangement of leaves on a stem and the branching of trees.

By recognizing the real-world applications of sequences and series, you can gain a deeper appreciation for their importance and relevance.

More Practice Problems

To solidify your understanding, here are a few more practice problems:

1. Evaluate the product of the sequence: ${ \frac{2}{4} \cdot \frac{4}{6} \cdot \frac{6}{8} \cdot \dots \cdot \frac{98}{100} }$.
2. Find the sum of the series: ${ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} }$.
3. Determine the limit of the sequence: ${ a_n = \frac{n^2 + 1}{2n^2 + 3n - 1} }$ as ${ n }$ approaches infinity.

Work through these problems, and don't hesitate to seek help or consult resources if you get stuck. Happy problem-solving!