Solving For L In The Series 1/3.6 + 1/6.9 + 1/9.12 + ... + 1/30.33
Hey guys! Today, we're diving into a fun math problem where we need to find the value of the series: L = 1/3.6 + 1/6.9 + 1/9.12 + ... + 1/30.33. This might look a bit daunting at first, but don't worry, we'll break it down step by step. Let's get started!
Understanding the Series
First, let's take a closer look at the series to see if we can spot any patterns. The series is: L = 1/(36) + 1/(69) + 1/(912) + ... + 1/(3033). Notice that each term has a similar structure: it's a fraction where the numerator is 1, and the denominator is a product of two numbers. Specifically, the numbers in the denominator are multiples of 3. The first number in each product increases by 3 each time (3, 6, 9, ..., 30), and the second number also increases by 3 each time (6, 9, 12, ..., 33).
Now, recognizing this pattern is the first key step. We can rewrite the series in a more general form. If we let n represent the term number (starting from 1), the nth term can be written as 1/((3n) * (3n + 3)). This general form helps us see the underlying structure more clearly. Another way to put this is that the nth term in the series can be represented as: 1 / (3n * (3n + 3)).
Let's try plugging in a few values of n to see if it matches the series. When n = 1, the term is 1/(3 * 6), which is the first term in the series. When n = 2, the term is 1/(6 * 9), which is the second term. When n = 10, the term is 1/(30 * 33), which is the last term in the series. So, this general form correctly represents our series. This is crucial because it allows us to use mathematical tools and techniques to simplify and solve the problem.
Partial Fractions: The Key Technique
The next trick up our sleeve is using partial fraction decomposition. This technique is super helpful when dealing with fractions where the denominator can be factored into simpler terms. In our case, we can rewrite each term in the series using partial fractions. The general idea behind partial fraction decomposition is to express a complex fraction as a sum or difference of simpler fractions.
For the nth term, 1/(3n * (3n + 3)), we want to find constants A and B such that: 1/(3n * (3n + 3)) = A/(3n) + B/(3n + 3). To find A and B, we need to clear the denominators. Multiply both sides of the equation by 3n * (3n + 3): 1 = A(3n + 3) + B(3n). Now, we can solve for A and B by choosing convenient values of n.
If we let n = 0, we get: 1 = A(30 + 3) + B(30), which simplifies to 1 = 3A. Therefore, A = 1/3. If we let n = -1, we get: 1 = A(3*(-1) + 3) + B(3*(-1)), which simplifies to 1 = -3B. Therefore, B = -1/3. So, we've found our constants! We can rewrite the nth term as: 1/(3n * (3n + 3)) = (1/3) * (1/(3n) - 1/(3n + 3)).
By applying this partial fraction decomposition, we've transformed each term into a difference of two simpler fractions, which is going to make the summation much easier to handle. This is the core insight that unlocks the solution to the problem. It's like turning a complicated puzzle into something more manageable by breaking it down into smaller pieces. The use of partial fractions is a classic technique in calculus and series problems, and mastering it can significantly improve your problem-solving skills.
Applying Partial Fractions to the Series
Now that we know how to decompose each term using partial fractions, let's apply this to the entire series. Remember, we found that 1/(3n * (3n + 3)) = (1/3) * (1/(3n) - 1/(3n + 3)). So, let's rewrite each term in the series L using this decomposition:
L = (1/3) * (1/3 - 1/6) + (1/3) * (1/6 - 1/9) + (1/3) * (1/9 - 1/12) + ... + (1/3) * (1/30 - 1/33). Notice that the (1/3) factor is common to every term, so we can factor it out: L = (1/3) * [(1/3 - 1/6) + (1/6 - 1/9) + (1/9 - 1/12) + ... + (1/30 - 1/33)].
The magic of this decomposition lies in what happens next. Look closely at the terms inside the brackets. You'll see a beautiful pattern of cancellation. The -1/6 in the first term cancels out with the +1/6 in the second term. Similarly, the -1/9 in the second term cancels out with the +1/9 in the third term. This pattern continues throughout the series. This type of cancellation is called a telescoping sum, and it's a powerful technique for simplifying series.
As you can imagine, most of the terms will cancel out, leaving only the first term (1/3) and the last term (-1/33). This is the essence of a telescoping sum – the series collapses like a telescope, leaving only the end terms. So, the series simplifies to: L = (1/3) * (1/3 - 1/33).
Now, we just need to do a little bit of arithmetic to finish the job. To subtract the fractions, we need a common denominator. The least common multiple of 3 and 33 is 33. So, we rewrite 1/3 as 11/33: L = (1/3) * (11/33 - 1/33) = (1/3) * (10/33). Finally, we multiply the fractions: L = 10/99. And there we have it! The value of the series L is 10/99.
Conclusion: The Answer and Key Takeaways
So, after carefully breaking down the series, using partial fractions, and recognizing the telescoping pattern, we've found that the value of L = 1/3.6 + 1/6.9 + 1/9.12 + ... + 1/30.33 is 10/99. The correct answer is A) 10/99.
This problem beautifully illustrates several important mathematical concepts:
- Pattern Recognition: The ability to spot patterns in a series is crucial for finding a solution. In this case, recognizing that the denominators were products of multiples of 3 was the first key step.
- Partial Fraction Decomposition: This technique allows us to rewrite complex fractions as a sum or difference of simpler fractions, making them easier to work with.
- Telescoping Sums: Recognizing telescoping sums can dramatically simplify series where most terms cancel out, leaving only the first and last terms.
By mastering these concepts, you'll be well-equipped to tackle a wide range of series problems. Remember, math is like a puzzle – each problem is a new challenge waiting to be solved. Keep practicing, and you'll become a master problem-solver in no time! Guys, keep up the great work, and see you in the next problem!