Use mathematical induction to prove that 2^(3n) - 3^n is divisible by 5 for all positive integers.
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check for n=1
2^3 - 3^1 = 8-3 = 5
assume for k:
2^(3k) - 3^k = 5m for some m
now plug in k+1
2^(3(k+1)) - 3^(k+1)
= 2^(3k+3) - 3^(k+1)
= 2^3 * 2^(2k) - 3*3^k
= 8*2^(3k) - 3*3^k
= 3*2^(3k) + 5*2^(3k) - 3*3^k
= 3(2^(3k) - 3^k) + 5*2^(3k)
= 3(5m) + 5*2^(3k)
which is a multiple of 5.
To prove that 2^(3n) - 3^n is divisible by 5 for all positive integers, we will use mathematical induction.
Step 1: Base Case
Let's start with the base case when n = 1.
When n = 1, we have 2^(3*1) - 3^1 = 8 - 3 = 5, which is divisible by 5.
Step 2: Inductive Hypothesis
Assume that for some positive integer k, 2^(3k) - 3^k is divisible by 5.
Step 3: Inductive Step
We need to prove that if the statement holds for k, it also holds for k + 1.
Using the inductive hypothesis, we have:
2^(3(k+1)) - 3^(k+1) = 2^(3k+3) - 3^(k+1)
We can rewrite this expression as:
(2^3 * 2^k) - (3 * 3^k)
Simplifying further, we get:
(8 * 2^k) - (3 * 3^k)
We notice that we have a term 2^k in the first part, and a term 3^k in the second part.
Considering the properties of modular arithmetic, let's divide each term by 5.
(8 * 2^k) - (3 * 3^k) ≡ (3 * 2^k) - (3 * 3^k) (mod 5)
Now, we can factor out 3 from both terms:
(3 * 2^k) - (3 * 3^k) ≡ 3 * (2^k - 3^k) (mod 5)
By the inductive hypothesis, we know that 2^k - 3^k is divisible by 5.
Let's denote it as m, where m is an integer.
Therefore, 3 * (2^k - 3^k) is divisible by 5.
This completes the inductive step.
Step 4: Conclusion
By mathematical induction, we have shown that for any positive integer n, 2^(3n) - 3^n is divisible by 5. This completes the proof.
To prove that 2^(3n) - 3^n is divisible by 5 for all positive integers using mathematical induction, we will follow a two-step process:
Step 1: Show that the statement holds true for the base case.
Step 2: Assume that the statement holds true for an arbitrary positive integer k and prove that it also holds true for k+1.
Step 1: Base case (n = 1)
When n = 1, we have 2^(3*1) - 3^1 = 2^3 - 3 = 8 - 3 = 5. Since 5 is divisible by 5, the statement holds true for the base case.
Step 2: Inductive step
Assume that the statement holds true for an arbitrary positive integer k. That is,
2^(3k) - 3^k is divisible by 5.
Now, we need to prove that the statement also holds true for k+1, which means we need to show that
2^(3(k+1)) - 3^(k+1) is divisible by 5.
Starting with the left-hand side:
2^(3(k+1)) - 3^(k+1)
= 2^(3k + 3) - 3^(k+1)
= 2^(3k) * 2^3 - 3 * 3^k
= 8 * 2^(3k) - 3 * 3^k
= (5 + 3) * 2^(3k) - 3 * 3^k
= 5 * 2^(3k) + 3 * 2^(3k) - 3 * 3^k
Now, looking at each term:
1. 5 * 2^(3k) is divisible by 5 because it has a factor of 5.
2. 3 * 2^(3k) can be rewritten as 2 * (2^(3k)), which is divisible by 5 due to the inductive hypothesis.
3. -3 * 3^k is also divisible by 5 because it has a factor of 5.
Since all three terms are divisible by 5, their sum, 2^(3(k+1)) - 3^(k+1), is also divisible by 5.
Hence, by mathematical induction, we have proven that 2^(3n) - 3^n is divisible by 5 for all positive integers.