Unlocking The Decimal Secrets Of √(4n²+13n+10) A Mathematical Journey

by Kenji Nakamura 70 views

Hey guys! Ever stumbled upon a math problem that just makes you scratch your head and go, "Hmm, that's a tricky one"? Well, I recently encountered a fascinating problem that I'm super excited to share with you all. It involves finding the first decimal place of a seemingly complex expression. Sounds intriguing, right? Let's dive in and unravel this mathematical mystery together!

The Challenge: Cracking the Code of √(4n²+13n+10)

The problem we're tackling today asks us to determine the first decimal place of the number a_n = √(4n²+13n+10) for every positive integer n. At first glance, this might seem like a daunting task. I mean, we're dealing with square roots, quadratic expressions, and an infinite set of numbers! But don't worry, we'll break it down step by step and conquer this challenge like true math detectives.

When I initially approached this problem, my first instinct was to try plugging in different values for n and see if any patterns emerged. It's a classic problem-solving technique, and sometimes, it can lead to unexpected breakthroughs. So, I started calculating the values of a_n for a few small values of n and meticulously noted down the first decimal place. This hands-on approach gave me a feel for how the expression behaves and hinted at a possible solution strategy. However, relying solely on numerical examples isn't enough to solve the problem definitively. We need a more rigorous approach to unveil the underlying mathematical structure.

Initial Explorations: Numerical Adventures and Pattern Recognition

Like any good math explorer, let's start our journey by plugging in some values and seeing what happens. When n equals 1, a_1 becomes the square root of (4 + 13 + 10), which simplifies to √27, approximately 5.196. So, the first decimal place is 1. For n equals 2, a_2 transforms into √(16 + 26 + 10), or √52, roughly 7.211, giving us a first decimal place of 2. Continuing this pattern, for n equals 3, a_3 equals √(36 + 39 + 10), which is √85, about 9.220, with a first decimal place of 2. These initial calculations provide us with a tangible sense of the problem and allow us to see the numbers in action. However, it's crucial to remember that observing a pattern for a few cases doesn't guarantee it holds true for all n. We need a more robust method to prove our conjecture.

The Quest for a Clever Squeeze: Bounding the Square Root

Our mission now is to find a way to squeeze our expression, √(4n²+13n+10), between two simpler expressions. This technique, often called bounding, is a powerful tool in mathematical analysis. If we can find two perfect squares that closely bound our expression, we can gain valuable insights into its decimal behavior. The key idea here is to find perfect squares that are slightly smaller and slightly larger than 4n²+13n+10. This will allow us to create upper and lower bounds for our square root expression.

So, let's think about perfect squares. We know that (2n)² equals 4n², which is close to our expression. But we need to account for the 13n + 10 part. Let's try (2n + k)², where k is some constant we need to figure out. Expanding this, we get 4n² + 4nk + k². Now, we want this to be close to 4n²+13n+10. If we set 4k approximately equal to 13, we get k around 3. Let's try k equals 2 and k equals 3 to see how they behave. We'll aim to manipulate our expression so that it fits snugly between two consecutive integers, thereby revealing its first decimal place.

The Art of Manipulation: Crafting the Perfect Inequalities

Now, let’s get our hands dirty with some algebraic manipulation. Our goal is to sandwich √(4n²+13n+10) between two consecutive integers. This will tell us which integer the square root is closest to and, consequently, its first decimal place. Let's start by comparing 4n²+13n+10 with (2n+3)². Expanding (2n+3)² gives us 4n²+12n+9. Notice that 4n²+13n+10 is slightly larger than 4n²+12n+9. This is a great start! It means that √(4n²+13n+10) is greater than 2n+3.

But we're not done yet. We need an upper bound as well. Let's try comparing 4n²+13n+10 with (2n+4)². Expanding (2n+4)² gives us 4n²+16n+16. Now, is 4n²+13n+10 smaller than this? We need to check the inequality: 4n²+13n+10 < 4n²+16n+16. Simplifying, we get 0 < 3n + 6, which is true for all positive integers n. This is fantastic news! We've successfully sandwiched our square root: 2n+3 < √(4n²+13n+10) < 2n+4. The real magic happens when we carefully choose our bounding squares. It's like finding the perfect fitting puzzle pieces that reveal the hidden picture.

Unveiling the Decimal Clues: A Closer Look at the Difference

We've made significant progress by bounding our expression between two consecutive integers. However, to pinpoint the first decimal place, we need to zoom in a little closer. We need to determine whether √(4n²+13n+10) is closer to 2n+3 or 2n+4. To do this, let's consider the midpoint between these two integers, which is 2n + 3.5. If √(4n²+13n+10) is less than 2n + 3.5, then its first decimal place will be less than 5. If it's greater, the first decimal place will be 5 or more. Squaring both sides of the inequality √(4n²+13n+10) < 2n + 3.5, we get 4n²+13n+10 < (2n + 3.5)². Expanding the right side, we have 4n²+13n+10 < 4n²+14n+12.25. Simplifying, we arrive at 0 < n + 2.25. This inequality holds true for all positive integers n. This tells us that √(4n²+13n+10) is always less than 2n + 3.5. The brilliance of this step lies in the strategic comparison with the midpoint. It allows us to make a decisive conclusion about the decimal behavior of our expression.

The Grand Finale: Decoding the First Decimal Place

We've reached the climax of our mathematical adventure! We've successfully shown that 2n+3 < √(4n²+13n+10) < 2n+3.5 for all positive integers n. This elegant inequality is the key to unlocking the mystery of the first decimal place. Since √(4n²+13n+10) is always greater than 2n+3 but less than 2n+3.5, its first decimal place must be either 1 or 2 or 3 or 4. It cannot be 5 or greater. Therefore, the first decimal place of a_n = √(4n²+13n+10) is either 1, 2, 3, or 4. This is a beautiful result! We started with a seemingly complex expression, navigated through a series of inequalities, and arrived at a clear and concise answer. The journey itself was as rewarding as the destination, showcasing the power of mathematical reasoning and the joy of discovery.

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