Calculating The Increase In Sum Of Products

Hey guys! Let's dive into a cool math problem. We're gonna figure out how much the sum of a series of products changes when we tweak those products a bit. Specifically, we're looking at the sum: A = 1 × 2 + 2 × 3 + 3 × 4 + … + 23 × 24. The core question is: what happens to the value of A if we increase each of the factors in each term by 2? Pretty neat, right?

So, the original sum (A) is made up of a bunch of terms. Each term is a product of two numbers. The first term is 1 times 2, the second is 2 times 3, and so on, all the way up to 23 times 24. We want to know how the entire sum changes when we bump up both numbers in each of these products. This kind of problem is a great way to practice our algebraic thinking and our understanding of how numbers work together. We'll explore this step by step, making sure everything is clear as mud. This is not just about getting the right answer; it's about understanding why that's the right answer. We'll use some basic algebraic principles, which will help us break down this problem, and help us see how the changes affect the overall outcome. This strategy will enable us to solve the problem and also give us a solid foundation for handling similar math questions. Let's get started. Believe me, it's easier than it looks. Ready to get our math hats on?

Breaking Down the Original Sum and the Modified Sum

Alright, let's start with the basics. We already know our original sum is: A = 1 × 2 + 2 × 3 + 3 × 4 + … + 23 × 24. This can also be written as a sigma notation which is the short way to say we're adding up a bunch of similar things. In this case, each term is in the form of n × (n + 1), and we are doing this for all the numbers from 1 to 23. It helps to understand the structure of the sum because it makes it way easier to manipulate.

Now, let's consider what happens when we increase each factor by 2. So, instead of 1 × 2, we have (1 + 2) × (2 + 2), and instead of 2 × 3, we have (2 + 2) × (3 + 2). In the end, each original term n × (n + 1) turns into (n + 2) × (n + 3). To make things clear, let's call this new sum A'. Therefore, A' will look like: (1 + 2) × (2 + 2) + (2 + 2) × (3 + 2) + (3 + 2) × (4 + 2) + … + (23 + 2) × (24 + 2). That is, (3 × 4) + (4 × 5) + (5 × 6) + … + (25 × 26). Our goal is to figure out the difference between A' and A. That will tell us how much the sum increased. The difference is the key here. Now that we've set up the problem and clarified the steps, we can move forward. We now know that our new sum uses the same number of terms as the old one, but it's the factors in each of those terms that are really changing. The beauty of math is how a few simple adjustments can completely change the final answer. We need to remember this as we move forward.

Calculating the Difference Between A' and A

Now, let's get down to the actual calculation. We need to find out how much larger A' is than A. It's a matter of finding the total value of each sum and then comparing them. However, we can use algebra to make our job a lot easier and avoid some number crunching. Remember, each term in A' is (n + 2) × (n + 3), and each term in A is n × (n + 1). Now, let's expand the terms in A': (n + 2) × (n + 3) = n² + 3n + 2n + 6 = n² + 5n + 6. If we apply the same principle to A, we already know each term is n × (n + 1) = n² + n.

So, when we find the difference between a term in A' and the corresponding term in A, we have: (n² + 5n + 6) - (n² + n) = 4n + 6. This is huge! It means that for each individual term, the difference is 4n + 6. That might sound complicated, but it's not. Remember, we are trying to find the sum of all these differences. So, we need to add up 4n + 6 for n from 1 to 23. This becomes (4 × 1 + 6) + (4 × 2 + 6) + (4 × 3 + 6) + … + (4 × 23 + 6). That looks a little bit like a problem from elementary school. We can break this down further: we can actually split this up into two sums: the sum of 4n and the sum of 6 (repeated 23 times, one for each term). The sum of 6, repeated 23 times, is simply 6 × 23 = 138. Now we just need the sum of 4n. This can be written as 4 × (1 + 2 + 3 + … + 23). The sum of the first n natural numbers is given by the formula n × (n + 1) / 2. Therefore, the sum of numbers from 1 to 23 is (23 × 24) / 2 = 276. Multiplying this by 4, we get 4 × 276 = 1104. Finally, the total increase will be 1104 + 138 = 1242. This is the difference between A' and A. This algebraic trick makes it much simpler to solve without needing to calculate all the original products first. Math is all about these simple tricks.

Unveiling the Final Answer and Its Significance

So, the answer is: the value of A increases by 1242 when each factor in each term is increased by 2. Boom! That’s it! We’ve successfully figured out how much our sum increased. We’ve not just calculated a number; we've shown how mathematical principles work together. We’ve used algebraic manipulation to make a seemingly complex problem manageable. The key was breaking the problem into smaller parts and seeing how each part contributes to the whole. In doing this, we can easily see how each factor changes the overall sum. This method works because we are using the properties of addition and multiplication to systematically adjust the values. This approach is highly useful when tackling similar problems. It helps you to understand, not just the answer, but the structure of the problem. This not only solves the question at hand but also sets a solid foundation for future math challenges. This kind of problem-solving approach is crucial in all sorts of mathematical and real-world scenarios. Think about it: a seemingly minor change in the components of a complex system can result in some significant results. Therefore, understanding the impact of these changes is a valuable skill.

In conclusion, we've gone from a series of products to a single answer: an increase of 1242. This journey underscores the power of algebraic thinking and the beauty of mathematics. It’s also a reminder that seemingly complex problems can be solved with the right strategies and a bit of patience. Congrats, guys! You did it! Keep practicing, and you'll find that these kinds of math puzzles get easier and more fun over time. Keep those math hats on. The world of numbers is full of exciting things to discover.