Sum Of The Series 1 + 4 + 7 + 10 + 13 + ... + (3n - 2) Explained

Hey guys! Ever stumbled upon a series of numbers that seem to follow a pattern and wondered if there's a neat way to add them all up? Well, today we're diving deep into one such intriguing series: 1 + 4 + 7 + 10 + 13 + ... + (3n - 2). This isn't just a random collection of numbers; it's a classic example of an arithmetic series, and we're going to explore how to find its sum. Get ready to flex those mathematical muscles and uncover the secrets behind this series!

Understanding Arithmetic Series

Before we jump into solving the sum of our specific series, let's take a step back and understand what an arithmetic series actually is. In simple terms, an arithmetic series is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is often referred to as the common difference. Think of it like climbing stairs where each step has the same height. In our series, 1 + 4 + 7 + 10 + 13 + ... + (3n - 2), you can see that we're adding 3 to each term to get the next one (4 - 1 = 3, 7 - 4 = 3, and so on). So, the common difference here is 3. This consistent pattern is what makes arithmetic series predictable and allows us to calculate their sums using some handy formulas. Knowing this basic concept is the key to unlocking the problem we're tackling today. We'll use the properties of arithmetic series to develop a strategy for finding the sum, so stay with me as we break it down step by step. Understanding the underlying principles will not only help us solve this particular problem but also equip you with the knowledge to handle similar series in the future. So, let's keep these concepts in mind as we move forward and delve deeper into the solution.

Identifying the Key Components of the Series

Now that we've got a handle on what arithmetic series are, let's zoom in on our specific series: 1 + 4 + 7 + 10 + 13 + ... + (3n - 2). To find the sum, we need to identify some crucial components. First off, what's the first term? Looking at the series, it's clear that the first term, often denoted as 'a', is 1. This is our starting point. Next, we need the common difference, which we already figured out is 3. This is the constant value we add to each term to get the next. But what about the last term? That's where the expression (3n - 2) comes in. This is a general formula that represents any term in the series, where 'n' is the position of the term. For example, if we want to find the 5th term, we'd plug in n = 5, giving us (3 * 5 - 2) = 13, which matches our series. Finally, we need to know the number of terms in the series. This is where 'n' really shines. It tells us how many terms we're adding up. So, in our case, we have 'n' terms. Identifying these components – the first term (a = 1), the common difference (d = 3), the last term (3n - 2), and the number of terms (n) – is like gathering the ingredients for a recipe. Once we have them, we can use the right formula to cook up the sum of the series. Keep these components in mind as we move on to the next step, where we'll put them to work.

Applying the Formula for the Sum of an Arithmetic Series

Alright, guys, we've identified the key components of our arithmetic series, and now it's time for the magic formula! The sum of an arithmetic series, often denoted as S_n, can be calculated using the following formula: S_n = n/2 * [2a + (n - 1)d] or S_n = n/2 * [a + l] where:

• S_n is the sum of the first 'n' terms
• n is the number of terms
• a is the first term
• d is the common difference
• l is the last term

This formula might look a bit intimidating at first, but trust me, it's a powerful tool. It essentially takes the average of the first and last terms and multiplies it by the number of terms. Think of it as finding the area of a trapezoid, where the parallel sides are the first and last terms, and the height is related to the number of terms. Now, let's plug in the values we identified earlier into the first formula. We know that a = 1, d = 3, and the number of terms is 'n'. Substituting these values into the formula, we get:

S_n = n/2 * [2(1) + (n - 1)3]

Now, it's just a matter of simplifying the expression. We'll distribute the 3 and combine like terms to get a more manageable formula. This is where our algebraic skills come into play. We're not just blindly plugging in numbers; we're manipulating the formula to make it easier to calculate the sum. So, let's get our algebraic hats on and simplify this expression step by step. This formula is the key to unlocking the sum of our series, and by understanding how it works, we're building a solid foundation for tackling similar problems in the future.

Simplifying the Formula

Okay, let's roll up our sleeves and simplify the formula we obtained in the previous step: S_n = n/2 * [2(1) + (n - 1)3]. The first thing we can do is distribute the 3 inside the brackets: S_n = n/2 * [2 + 3n - 3]. Now, let's combine the constant terms inside the brackets: S_n = n/2 * [3n - 1]. We're getting closer! To further simplify, we can distribute the n/2 across the terms inside the brackets: S_n = (n/2) * (3n) - (n/2) * (1). This gives us: S_n = (3n^2)/2 - n/2. Now, to make it look even cleaner, we can combine these terms by finding a common denominator: S_n = (3n^2 - n) / 2. And there you have it! We've successfully simplified the formula. This elegant expression, S_n = (3n^2 - n) / 2, is the key to calculating the sum of our arithmetic series for any number of terms 'n'. It's a much more manageable form than the original formula, and it allows us to quickly find the sum without having to add up each term individually. This simplification process is a crucial part of problem-solving in mathematics. It's not just about plugging in numbers; it's about manipulating expressions to make them easier to work with. So, we've taken a potentially complex formula and transformed it into a simple, powerful tool. Now, let's see this tool in action as we calculate the sum for a specific example.

Calculating the Sum for a Specific Example

Now that we have our simplified formula, S_n = (3n^2 - n) / 2, let's put it to the test with a specific example. Let's say we want to find the sum of the first 10 terms of the series: 1 + 4 + 7 + 10 + 13 + ... To do this, we simply substitute n = 10 into our formula: S_10 = (3(10)^2 - 10) / 2. Now, let's calculate: S_10 = (3 * 100 - 10) / 2 S_10 = (300 - 10) / 2 S_10 = 290 / 2 S_10 = 145. So, the sum of the first 10 terms of the series is 145. Isn't that neat? We were able to find the sum without having to manually add up all 10 terms. Our simplified formula made the process quick and easy. This example demonstrates the power of mathematical formulas. They allow us to solve problems efficiently and accurately. Instead of brute-force calculation, we can use a formula to arrive at the answer with minimal effort. This is why understanding and being able to apply formulas is such an important skill in mathematics. But we're not stopping here. Let's take it a step further and discuss a different approach to solving this problem, just to see the versatility of mathematical methods. So, stick around as we explore another way to tackle this arithmetic series.

Alternative Approach: Using the Average of the First and Last Term

Alright, math enthusiasts, let's explore another cool method to find the sum of our arithmetic series. Remember, we said the formula S_n = n/2 * [2a + (n - 1)d] can also be expressed as S_n = n/2 * [a + l], where 'l' is the last term? This second formula offers a slightly different perspective. It tells us that the sum of an arithmetic series is simply the average of the first term ('a') and the last term ('l'), multiplied by the number of terms ('n'). It's like finding the average value of all the terms and then multiplying it by the number of terms. To use this formula, we need to find the last term, which is given by 3n - 2. So, if we want to find the sum of the first 'n' terms, the last term is simply 3n - 2. Now, let's plug 'a' and 'l' into our formula: S_n = n/2 * [1 + (3n - 2)]. Simplifying inside the brackets, we get: S_n = n/2 * [3n - 1]. Notice something familiar? This is the same expression we had before we distributed the n/2 in our previous method! This is a great confirmation that our previous simplification was correct. Now, distributing the n/2, we get: S_n = (3n^2)/2 - n/2, which simplifies to: S_n = (3n^2 - n) / 2. Voila! We arrived at the same simplified formula using a different approach. This demonstrates the beauty of mathematics – there are often multiple paths to the same destination. Using the average of the first and last term provides an intuitive way to understand the sum of an arithmetic series. It highlights the symmetry and patterns inherent in these sequences. So, we've now seen two different ways to solve this problem, reinforcing our understanding and giving us more tools in our mathematical toolkit. But before we wrap things up, let's recap the key takeaways and solidify our understanding.

Key Takeaways and Conclusion

Okay, guys, let's recap what we've learned today! We embarked on a journey to find the sum of the arithmetic series 1 + 4 + 7 + 10 + 13 + ... + (3n - 2), and we've uncovered some valuable insights along the way. First, we defined what an arithmetic series is – a sequence where the difference between consecutive terms is constant. Then, we identified the key components of our series: the first term (a = 1), the common difference (d = 3), and the last term (3n - 2). Next, we applied the formula for the sum of an arithmetic series, S_n = n/2 * [2a + (n - 1)d], and simplified it to a more manageable form: S_n = (3n^2 - n) / 2. We even saw how this formula can be derived from S_n = n/2 * [a + l]. We then put our formula to the test with a specific example, finding the sum of the first 10 terms to be 145. Finally, we explored an alternative approach using the average of the first and last term, demonstrating the versatility of mathematical methods. So, what are the key takeaways? We've learned that arithmetic series have predictable patterns that allow us to calculate their sums using formulas. We've also seen the importance of simplifying expressions to make them easier to work with. And we've discovered that there are often multiple ways to approach a mathematical problem. This journey through the world of arithmetic series has not only given us a specific solution but also equipped us with valuable problem-solving skills that can be applied to a wide range of mathematical challenges. So, keep exploring, keep questioning, and keep those mathematical gears turning! You've got this!