Proof The Remainder Of N² + (n + 1)² + (n + 2)² Divided By 3 Is Always 2

Hey guys! Ever stumbled upon a math problem that seems like a puzzle? Well, let's dive into one today! We're going to prove that for any natural number n greater than or equal to 1, the number n² + (n + 1)² + (n + 2)² always leaves a remainder of 2 when divided by 3. Sounds interesting, right? Let's break it down step-by-step!

Understanding the Problem

Before we jump into the proof, let's make sure we understand what the problem is asking. We're dealing with natural numbers, which are the counting numbers (1, 2, 3, and so on). We're given an expression n² + (n + 1)² + (n + 2)², and we need to show that no matter what natural number we plug in for n, the result will always have a remainder of 2 when divided by 3.

Why is this important? This kind of problem highlights the beauty of mathematical proofs. It's not just about finding an answer for one specific case; it's about showing that a statement is true for all cases within a given set. This is the power of mathematical reasoning!

Our Game Plan: To tackle this, we'll use a bit of algebra and some clever thinking about remainders. We'll expand the expression, simplify it, and then look for patterns that relate to divisibility by 3. Think of it as detective work, but with numbers!

Expanding and Simplifying the Expression

The first step in our mathematical journey is to expand and simplify the expression n² + (n + 1)² + (n + 2)². This will help us to better understand its structure and how it behaves. Let's start by expanding the squared terms:

• (n + 1)² = n² + 2n + 1
• (n + 2)² = n² + 4n + 4

Now, let's substitute these expansions back into our original expression:

• n² + (n + 1)² + (n + 2)² = n² + (n² + 2n + 1) + (n² + 4n + 4)

Next, we combine like terms to simplify the expression:

• n² + n² + 2n + 1 + n² + 4n + 4 = 3n² + 6n + 5

So, our simplified expression is 3n² + 6n + 5. This looks much cleaner and easier to work with. Notice that we have terms that are multiples of 3 (3n² and 6n), which is a good sign since we're interested in divisibility by 3. But we also have that pesky +5 at the end. How do we deal with that?

Focusing on Divisibility by 3

Okay, so we've simplified our expression to 3n² + 6n + 5. The key to this problem lies in understanding how each part of this expression behaves when divided by 3. Notice that 3n² is always divisible by 3, right? Because it's 3 multiplied by something. Similarly, 6n is also always divisible by 3. That's awesome! It means these terms won't contribute to the remainder when we divide the whole expression by 3.

But what about the +5? Well, we can think of 5 as 3 + 2. Now our expression looks like this:

• 3n² + 6n + 5 = 3n² + 6n + 3 + 2

Aha! We've separated out another multiple of 3! Now we have 3n² + 6n + 3, which is clearly divisible by 3. And we're left with that lone +2. This is the magic ingredient!

We can rewrite the expression as:

• 3n² + 6n + 3 + 2 = 3(n² + 2n + 1) + 2

See what we did there? We factored out a 3 from the first three terms. This clearly shows that the entire first part of the expression, 3(n² + 2n + 1), is divisible by 3. So, the only thing that can affect the remainder when we divide by 3 is that +2 at the end. And that, my friends, is exactly what we wanted to prove!

Formalizing the Proof

We've walked through the logic, and we've seen why the remainder is always 2. But in mathematics, we need to write things down formally to make our argument airtight. So, let's put it all together in a concise proof.

Proof:

Let n be a natural number such that n ≥ 1. Consider the expression n² + (n + 1)² + (n + 2)². We want to show that this expression leaves a remainder of 2 when divided by 3.

First, expand and simplify the expression:

• n² + (n + 1)² + (n + 2)² = n² + (n² + 2n + 1) + (n² + 4n + 4)
• • = 3n² + 6n + 5*

Next, rewrite the constant term 5 as 3 + 2:

• 3n² + 6n + 5 = 3n² + 6n + 3 + 2

Now, factor out a 3 from the first three terms:

• 3n² + 6n + 3 + 2 = 3(n² + 2n + 1) + 2

Let k = n² + 2n + 1. Then, the expression becomes:

• 3k + 2

Since k is an integer (because n is a natural number), 3k is divisible by 3. Therefore, the expression 3k + 2 leaves a remainder of 2 when divided by 3. This completes the proof.

Conclusion

So, there you have it! We've successfully proven that for any natural number n ≥ 1, the number n² + (n + 1)² + (n + 2)² always leaves a remainder of 2 when divided by 3. How cool is that? We took a seemingly abstract problem and broke it down using algebra and logical reasoning. This is what makes math so fascinating – the ability to uncover patterns and prove truths that hold for an infinite number of cases.

Remember, guys, the next time you encounter a challenging math problem, don't be afraid to dive in and explore. Break it down, simplify it, and look for the underlying patterns. You might just surprise yourself with what you can discover!

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