Divisibility Proof: $3^{2n+1} + 2^{n+2}$ Divisible By 7

Hey guys! Let's dive into an exciting mathematical problem today. We're going to prove that for all natural numbers n, the expression $3^{2n+1} + 2^{n+2}$ is divisible by 7. This is a classic example of a problem that can be elegantly solved using mathematical induction. So, buckle up and let's get started!

Understanding the Problem

Before we jump into the proof, let’s make sure we fully understand what we're trying to show. The statement says that if you take any natural number (n = 0, 1, 2, 3, ...), plug it into the expression $3^{2n+1} + 2^{n+2}$, the result will always be a multiple of 7. For instance, if n = 0, we have $3^{2(0)+1} + 2^{0+2} = 3^1 + 2^2 = 3 + 4 = 7$, which is indeed divisible by 7. If n = 1, we get $3^{2(1)+1} + 2^{1+2} = 3^3 + 2^3 = 27 + 8 = 35$, which is also divisible by 7. But these examples don't prove it for all natural numbers. That's where mathematical induction comes in handy.

Mathematical induction is a powerful technique used to prove statements that hold for all natural numbers. It’s like setting up a line of dominoes – if you can knock over the first domino and show that each domino will knock over the next one, then you know all the dominoes will fall. In our case, the dominoes are the natural numbers, and we want to show that the statement "$3^{2n+1} + 2^{n+2}$ is divisible by 7" holds for each of them.

The Principle of Mathematical Induction

Mathematical induction involves three main steps:

1. Base Case: Show that the statement is true for the smallest natural number, usually n = 0 or n = 1.
2. Inductive Hypothesis: Assume that the statement is true for some arbitrary natural number k. This means we assume $3^{2k+1} + 2^{k+2}$ is divisible by 7.
3. Inductive Step: Prove that if the statement is true for k, then it must also be true for k + 1. This means we need to show that $3^{2(k+1)+1} + 2^{(k+1)+2}$ is divisible by 7, given that $3^{2k+1} + 2^{k+2}$ is divisible by 7.

Once we complete these three steps, the principle of mathematical induction guarantees that the statement is true for all natural numbers.

Proof by Mathematical Induction

Now, let's apply these steps to our problem.

1. Base Case

We'll start with the base case n = 0. We need to show that $3^{2(0)+1} + 2^{0+2}$ is divisible by 7.

$3^{2(0)+1} + 2^{0+2} = 3^1 + 2^2 = 3 + 4 = 7$

Since 7 is divisible by 7, the base case holds true.

2. Inductive Hypothesis

Next, we assume that the statement is true for some arbitrary natural number k. This means we assume that $3^{2k+1} + 2^{k+2}$ is divisible by 7. In other words, we assume there exists an integer m such that:

$3^{2k+1} + 2^{k+2} = 7m$

This is our inductive hypothesis, and it's the crucial assumption we'll use in the next step.

3. Inductive Step

Now, we need to prove that if the statement is true for k, then it must also be true for k + 1. That is, we need to show that $3^{2(k+1)+1} + 2^{(k+1)+2}$ is divisible by 7. Let's expand this expression:

$3^{2(k+1)+1} + 2^{(k+1)+2} = 3^{2k+3} + 2^{k+3}$

Our goal is to manipulate this expression to show that it's a multiple of 7, using our inductive hypothesis. Let's rewrite the terms:

$3^{2k+3} + 2^{k+3} = 3^{2k+1+2} + 2^{k+2+1} = 3^{2k+1} imes 3^2 + 2^{k+2} imes 2^1 = 9 imes 3^{2k+1} + 2 imes 2^{k+2}$

Now, we want to somehow use our inductive hypothesis ($3^{2k+1} + 2^{k+2} = 7m$) to rewrite this expression. Notice that we have $3^{2k+1}$ and $2^{k+2}$ terms. Let's try to isolate one of these terms in our inductive hypothesis:

$3^{2k+1} = 7m - 2^{k+2}$

Now, substitute this back into our expression:

$9 imes 3^{2k+1} + 2 imes 2^{k+2} = 9(7m - 2^{k+2}) + 2 imes 2^{k+2} = 63m - 9 imes 2^{k+2} + 2 imes 2^{k+2}$

Combine the $2^{k+2}$ terms:

$63m - 9 imes 2^{k+2} + 2 imes 2^{k+2} = 63m - 7 imes 2^{k+2}$

Now, we can factor out a 7:

$63m - 7 imes 2^{k+2} = 7(9m - 2^{k+2})$

Since $9m - 2^{k+2}$ is an integer (because m and k are integers), we have shown that $3^{2(k+1)+1} + 2^{(k+1)+2}$ is a multiple of 7. This completes the inductive step.

Conclusion

We have successfully shown:

1. Base Case: The statement is true for n = 0.
2. Inductive Hypothesis: We assumed the statement is true for some arbitrary natural number k.
3. Inductive Step: We proved that if the statement is true for k, then it is also true for k + 1.

Therefore, by the principle of mathematical induction, the statement “For all natural numbers n, $3^{2n+1} + 2^{n+2}$ is divisible by 7” is true. Awesome job, guys! We tackled a cool math problem and learned how to use mathematical induction. Keep practicing, and you'll become math wizards in no time!

This proof demonstrates the power of mathematical induction in proving statements that hold for an infinite number of cases. It's a fundamental technique in discrete mathematics and computer science, and mastering it will open doors to solving many more challenging problems. So, keep exploring, keep learning, and most importantly, keep having fun with math!

Further Exploration

If you're interested in diving deeper into mathematical induction and divisibility proofs, here are a few topics you might want to explore:

• Strong Induction: A variation of mathematical induction where you assume the statement is true for all natural numbers up to k, not just k itself.
• Divisibility Rules: Learn other divisibility rules for different numbers and try to prove them using mathematical induction.
• Recursive Definitions: Explore how mathematical induction is used to prove properties of recursively defined sequences and functions.

By exploring these topics, you'll gain a deeper understanding of mathematical induction and its applications in various areas of mathematics and computer science. Happy learning! Remember, the key to mastering any mathematical concept is practice, practice, practice. So, don't be afraid to tackle challenging problems and push your boundaries. You've got this!