Calculate ∑1/((4k-1)(4k+4)): A Step-by-Step Guide

Hey there, math enthusiasts! Ever stumbled upon a seemingly daunting infinite series and felt a mix of curiosity and intimidation? Well, you're not alone! Infinite series, like the one we're tackling today – ∑1/((4k-1)(4k+4)) – can appear complex at first glance. But fear not, because with the right tools and a dash of mathematical ingenuity, we can unravel their mysteries and arrive at elegant solutions. So, grab your thinking caps, and let's dive into the fascinating world of telescopic sums and partial fraction decomposition to conquer this challenge!

Understanding the Challenge: The Series ∑1/((4k-1)(4k+4))

Before we jump into the solution, let's take a moment to appreciate the series we're about to tackle. The series ∑1/((4k-1)(4k+4)) represents the sum of an infinite number of terms, where each term is a fraction determined by the value of 'k'. As 'k' increases from 1 to infinity, the terms of the series become progressively smaller. But here's the million-dollar question: does this infinite sum converge to a finite value, or does it grow without bound? This is where the magic of telescopic sums comes into play, allowing us to transform the series into a more manageable form. To truly grasp the concept, we need to first understand the building blocks of this series and how each term contributes to the overall sum. It's like understanding the individual notes in a melody before appreciating the entire symphony. This initial understanding is crucial for choosing the right approach and avoiding common pitfalls in series calculations. Remember, a clear problem definition is half the solution!

The Power of Telescopic Sums: A Strategic Approach

The key to cracking this problem lies in the strategic use of telescopic sums. Telescopic sums are a brilliant technique for evaluating series where intermediate terms cancel out, leaving behind a simplified expression. Think of it like a collapsing telescope – the internal sections slide in, leaving only the first and last sections visible. In our case, we'll decompose the fraction 1/((4k-1)(4k+4)) into partial fractions, which will then allow for this beautiful cancellation to occur. This approach isn't just a clever trick; it's a powerful method with wide applications in calculus and analysis. Recognizing when a series can be expressed as a telescopic sum is a crucial skill for any aspiring mathematician. It often involves looking for patterns and relationships between consecutive terms. By mastering this technique, you'll be equipped to tackle a whole new class of series problems with confidence and elegance. So, let's roll up our sleeves and get ready to witness the magic of cancellation in action!

Unveiling the Magic: Partial Fraction Decomposition

Our first step towards unlocking the telescopic sum is to decompose the fraction 1/((4k-1)(4k+4)) into partial fractions. This technique allows us to express a complex fraction as a sum of simpler fractions, which are easier to work with. The idea is to find constants A and B such that: 1/((4k-1)(4k+4)) = A/(4k-1) + B/(4k+4). To find A and B, we'll multiply both sides by (4k-1)(4k+4), which gives us: 1 = A(4k+4) + B(4k-1). Now, we can strategically choose values of 'k' to eliminate one variable and solve for the other. For instance, setting k = 1/4 eliminates B, allowing us to solve for A. Similarly, setting k = -1 eliminates A, allowing us to solve for B. Once we've found A and B, we'll have successfully decomposed the fraction into its partial fractions, paving the way for the telescopic sum to reveal itself. This process is a cornerstone of many calculus techniques and is well-worth mastering. It's like having a secret decoder ring for complex fractions!

Finding the Constants: Solving for A and B

Let's put our partial fraction decomposition plan into action and solve for the constants A and B. Remember our equation: 1 = A(4k+4) + B(4k-1). To find A, we'll set k = 1/4. This makes the term with B equal to zero, leaving us with: 1 = A(4(1/4) + 4) = A(1 + 4) = 5A. Solving for A, we get A = 1/5. Now, to find B, we'll set k = -1. This makes the term with A equal to zero, leaving us with: 1 = B(4(-1) - 1) = B(-4 - 1) = -5B. Solving for B, we get B = -1/5. So, we've successfully found our constants: A = 1/5 and B = -1/5. This means we can rewrite our fraction as: 1/((4k-1)(4k+4)) = (1/5)/(4k-1) - (1/5)/(4k+4). Notice how we've transformed a single complex fraction into the difference of two simpler fractions. This is the key step that allows the telescopic sum to work its magic. With A and B in hand, we're one step closer to unlocking the solution to our infinite series!

The Telescopic Collapse: Witnessing the Cancellation

With our partial fraction decomposition complete, we can now rewrite our series as: ∑[ (1/5)/(4k-1) - (1/5)/(4k+4) ] from k=1 to infinity. Let's expand the first few terms of this series to see the cancellation pattern: (1/5)(1/3 - 1/8) + (1/5)(1/7 - 1/12) + (1/5)(1/11 - 1/16) + ... Notice how the terms start to cancel out! The -1/8 from the first term will cancel with a +1/8 term later in the series (though it's not immediately obvious in the first few terms, you'll see it emerge as you write out more terms). This is the essence of a telescopic sum – intermediate terms gracefully eliminate each other. To make this cancellation more apparent, let's write out the partial sum, Sn, which is the sum of the first 'n' terms: Sn = (1/5) [ (1/3 - 1/8) + (1/7 - 1/12) + (1/11 - 1/16) + ... + (1/(4n-1) - 1/(4n+4)) ]. As 'n' approaches infinity, the terms 1/(4n-1) and 1/(4n+4) will approach zero. However, not all terms will cancel completely. A few initial terms will remain, giving us a finite value for the sum. This cancellation dance is what makes telescopic sums so powerful and elegant. It's like watching a carefully choreographed routine where everything falls into place perfectly!

Finding the Limit: Summing to Infinity

As we observed in the previous step, the series telescopes beautifully, with most terms canceling out. To find the final sum, we need to determine which terms survive the cancellation as 'n' approaches infinity. Let's revisit our partial sum: Sn = (1/5) [ (1/3 - 1/8) + (1/7 - 1/12) + (1/11 - 1/16) + ... + (1/(4n-1) - 1/(4n+4)) ]. Writing out more terms, we see that the terms 1/3, 1/7, 1/11 remain. It is a series of the form 1/(4k-1). The number of these terms is such that 4k-1 < 4n+4 => k < n+ 5/4, which means that there are going to be floor(n+ 5/4) such terms. But in fact, we also need to consider the negative terms. For these, the condition is 4k+4 < 4n+4, or simply k < n. Thus, we have that the limit when n goes to infinity of the sum is equal to

(1/5) * (1/3 + 1/7 + 1/11 + 1/15 ) = 1/5 * ( (1/3 + 1/7) + (1/11 + 1/15) ) = 1/5 * ( 10/21 + 26/165 ) = 1/5 * ( 10/21 + 26/165 ) = 1/5 * ( (1055 + 267) / 1155 ) = 1/5 * ( 550 + 182 ) / 1155 = 1/5 * 732 / 1155 = 732 / 5775 = 244 / 1925

Thus, the sum of the infinite series ∑1/((4k-1)(4k+4)) is 244/1925. This elegant result showcases the power of telescopic sums and partial fraction decomposition. By breaking down the complex fraction into simpler components and observing the cancellation pattern, we were able to arrive at a concise and satisfying solution. This journey through infinite series highlights the beauty and interconnectedness of mathematical concepts. It's a testament to the fact that even seemingly daunting problems can be conquered with the right tools and a persistent spirit. So, the next time you encounter an infinite series, remember the power of telescopic sums and partial fraction decomposition – they might just be the keys to unlocking its secrets!

Real-World Applications: Beyond the Textbook

While our exploration of ∑1/((4k-1)(4k+4)) might seem purely academic, the techniques we've employed have far-reaching applications in various fields. Telescopic sums, in particular, pop up in areas like physics, engineering, and computer science. For example, they can be used to model systems where quantities accumulate or dissipate over time, with intermediate terms canceling out to reveal the overall change. In finance, telescopic sums can help analyze investment strategies where gains and losses in different periods offset each other. The concept of partial fraction decomposition, on the other hand, is fundamental in calculus for integrating rational functions. It's also used in electrical engineering to analyze circuits and in chemistry to study reaction kinetics. So, the skills we've honed in solving this seemingly abstract series problem are actually valuable tools for tackling real-world challenges. This underscores the importance of understanding fundamental mathematical concepts – they provide the foundation for solving a wide range of practical problems. It's like learning the alphabet – once you've mastered the basics, you can write entire stories!

Conclusion: A Triumph of Mathematical Ingenuity

Our journey through the calculation of ∑1/((4k-1)(4k+4)) has been a testament to the power of mathematical thinking. We started with a seemingly complex infinite series and, by employing the techniques of telescopic sums and partial fraction decomposition, we were able to arrive at a beautiful and concise solution: 244/1925. This process not only demonstrates the elegance of mathematics but also highlights the importance of strategic problem-solving. By breaking down the problem into smaller, manageable steps, we were able to navigate the intricacies of the series and reveal its hidden structure. The telescopic collapse, where intermediate terms cancel out, is a particularly satisfying illustration of mathematical harmony. And the fact that these techniques have broad applications in diverse fields underscores the practical value of mathematical knowledge. So, let's celebrate this triumph of ingenuity and continue to explore the fascinating world of mathematics with curiosity and enthusiasm. After all, every mathematical problem is an opportunity to learn, grow, and discover the hidden beauty that lies within!