Divisibility Proof: Mathematical Induction With 6^(2n) + 3^(n+2) + 3^n

Hey guys! Today, we're diving into a fascinating problem in algebra: proving that the expression 6^(2n) + 3^(n+2) + 3^n is divisible by 11 for all natural numbers n. We're going to tackle this using the powerful method of mathematical induction. So, buckle up, and let's get started!

Understanding Mathematical Induction

Before we jump into the specifics, let's quickly recap what mathematical induction is all about. Mathematical induction is a bulletproof technique for proving statements that hold true for all natural numbers (1, 2, 3, and so on). It's like setting up a chain reaction – if you can show the first domino falls, and that each domino knocks over the next, then you know all the dominos will fall.

The process involves three key steps:

1. Base Case: Show the statement is true for the smallest natural number (usually n = 1). This is our first domino.
2. Inductive Hypothesis: Assume the statement is true for some arbitrary natural number k. This is assuming a domino in the middle will fall.
3. Inductive Step: Prove the statement is true for the next natural number, k + 1, assuming it's true for k. This is showing that if a domino falls, it will knock over the next one.

Once we nail these three steps, we've proven the statement is true for all natural numbers. Pretty neat, huh?

Step 1: Base Case (n = 1)

Okay, let's get our hands dirty with our specific problem. The first thing we need to do is check if the expression holds true for n = 1. So, we'll substitute n = 1 into our expression:

6^(2*1) + 3^(1+2) + 3^1 = 6^2 + 3^3 + 3 = 36 + 27 + 3 = 66

Now, is 66 divisible by 11? Absolutely! 66 / 11 = 6, with no remainder. So, our base case holds strong. This is great news, it's our first domino falling perfectly.

Step 2: Inductive Hypothesis

Next up, we need to make our inductive hypothesis. This is where we assume the statement is true for some arbitrary natural number k. In other words, we're assuming that 6^(2k) + 3^(k+2) + 3^k is divisible by 11. This means we can write it as 11m, where m is some integer. So, let's state our assumption clearly:

Assume that 6^(2k) + 3^(k+2) + 3^k = 11m, for some integer m.

This is a crucial step. We're not proving anything yet; we're simply assuming the statement is true for k. Think of it as setting up the next domino in our chain, trusting it will fall if the previous one does.

Step 3: Inductive Step (Proving for n = k + 1)

Now for the main event: the inductive step! This is where we need to prove that if our statement is true for k, it must also be true for k + 1. In other words, we need to show that 6^(2(k+1)) + 3^((k+1)+2) + 3^(k+1) is divisible by 11, assuming our inductive hypothesis is true.

Let's start by writing out the expression for n = k + 1:

6^(2(k+1)) + 3^((k+1)+2) + 3^(k+1) = 6^(2k+2) + 3^(k+3) + 3^(k+1)

Our goal is to manipulate this expression until we can somehow use our inductive hypothesis (6^(2k) + 3^(k+2) + 3^k = 11m) to show it's a multiple of 11.

Let's break down the terms and rewrite them strategically:

• 6^(2k+2) = 6^(2k) * 6^2 = 36 * 6^(2k)
• 3^(k+3) = 3^(k+2) * 3^1 = 3 * 3^(k+2)
• 3^(k+1) = 3^k * 3^1 = 3 * 3^k

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

36 * 6^(2k) + 3 * 3^(k+2) + 3 * 3^k

This is where the magic happens! We want to somehow isolate the 6^(2k) + 3^(k+2) + 3^k term from our inductive hypothesis. To do this, we can rewrite the expression by cleverly adding and subtracting terms:

36 * 6^(2k) + 3 * 3^(k+2) + 3 * 3^k = 36 * 6^(2k) + 36 * 3^(k+2) + 36 * 3^k - 33 * 3^(k+2) - 33 * 3^k

Notice what we did there? We essentially multiplied our inductive hypothesis expression by 36 and then subtracted the extra terms to balance it out. Now we can factor out a 36 from the first three terms:

36 * (6^(2k) + 3^(k+2) + 3^k) - 33 * 3^(k+2) - 33 * 3^k

Aha! There's our inductive hypothesis term! We know from our assumption that 6^(2k) + 3^(k+2) + 3^k = 11m, so let's substitute that in:

36 * (11m) - 33 * 3^(k+2) - 33 * 3^k

Now, we can factor out a 33 from the last two terms:

36 * (11m) - 33 * (3^(k+2) + 3^k)

And finally, we can factor out an 11 from the entire expression:

11 * (36m - 3 * (3^(k+2) + 3^k))

Boom! We've done it! We've shown that 6^(2(k+1)) + 3^((k+1)+2) + 3^(k+1) can be written as 11 multiplied by some integer (36m - 3 * (3^(k+2) + 3^k)), which means it's divisible by 11. This is like seeing our domino knocking over the next one in the chain.

Conclusion: The Chain Reaction is Complete!

So, let's recap what we've achieved:

1. We showed the base case (n = 1) holds true.
2. We made the inductive hypothesis, assuming the statement is true for some k.
3. We proved the inductive step, showing that if it's true for k, it's also true for k + 1.

By successfully completing these three steps, we've proven, using the principle of mathematical induction, that the expression 6^(2n) + 3^(n+2) + 3^n is divisible by 11 for all natural numbers n.

This is a classic example of how mathematical induction can be used to prove statements about sequences and divisibility. It might seem a bit tricky at first, but with practice, you'll get the hang of it. Remember the key is to break down the problem into smaller, manageable steps, and to use your inductive hypothesis as a powerful tool to connect the k case to the k + 1 case.

Keep exploring, keep questioning, and keep learning, guys! You've got this!