Prove the following statements by mathematical induction.
57 divides 7^(n + 2) + 8^(2n + 1), for n ∈ Z+
check for n=1
7^3 + 8^3 = 855 = 57*15
assume P(k), then check for n=k+1
7^(k+1+2) + 8^(2(k+1)+1)
= 7 * 7^(k+2) + 8^2 * 8^(2k+1)
= 7 (7^(k+2) + 8^(2k+1)) + 57 * 8^(2k+1)
both terms are multiples of 57, so P(k+1)
QED
To prove the statement "57 divides 7^(n + 2) + 8^(2n + 1)" by mathematical induction, we need to show that it holds true for the base case and then prove the induction step.
**Base Case:**
Let's first consider n = 1.
Substituting n = 1 into the expression, we have 7^(1 + 2) + 8^(2(1) + 1) = 7^3 + 8^3 = 343 + 512 = 855.
We need to show that 57 divides 855.
Dividing 855 by 57 gives us 855 ÷ 57 = 15.
Since 855 is divisible by 57 with no remainder, the base case holds.
**Inductive Step:**
Now, let's assume that for some positive integer k, the statement holds true, i.e., 57 divides 7^(k + 2) + 8^(2k + 1).
We need to prove that it also holds true for k + 1, i.e., 57 divides 7^((k + 1) + 2) + 8^(2(k + 1) + 1).
Starting with the left-hand side (LHS) of the expression:
7^((k + 1) + 2) + 8^(2(k + 1) + 1)
Notice that we can rewrite this as 7^(k + 3) + 8^(2k + 3).
Using the properties of exponents, we can rewrite this further as 7^3 * 7^k + 8^3 * 8^(2k).
Now, let's apply the assumption we made earlier, where 57 divides 7^(k + 2) + 8^(2k + 1):
7^(k + 2) + 8^(2k + 1) = 57m (where m is an integer).
Substituting this into the expression:
7^3 * 7^k + 8^3 * 8^(2k) = (57m) * 7 + (57m) * 64.
Now, let's factor out 57 from both terms:
(57m) * (7 + 64).
Simplifying this expression gives us:
57 * 65m.
Since 65 is a multiple of 57 (65 = 57 + 8), we can rewrite this as:
57 * (1 + 8m).
Therefore, we have shown that for k + 1, the statement "57 divides 7^((k + 1) + 2) + 8^(2(k + 1) + 1)" is true.
Since we have proven the base case and the induction step, by mathematical induction, we have shown that for all positive integers n, 57 divides 7^(n + 2) + 8^(2n + 1).
To prove that 57 divides 7^(n + 2) + 8^(2n + 1) for all positive integers n using mathematical induction, we will follow the steps of the induction proof:
Step 1: Base Case
The base case is n = 1. We need to show that 57 divides 7^(1 + 2) + 8^(2 * 1 + 1).
Substituting n = 1, we have:
7^(1 + 2) + 8^(2 * 1 + 1) = 7^3 + 8^3 = 343 + 512 = 855
To prove that 57 divides 855, we divide 855 by 57 and observe that it gives us a quotient of 15 with no remainder.
Therefore, the base case holds true.
Step 2: Inductive Hypothesis
Assume that for some positive integer k, 57 divides 7^(k + 2) + 8^(2k + 1).
Step 3: Inductive Step
We need to prove that if the statement is true for k, it must also be true for k + 1.
This involves showing that if 57 divides 7^(k + 2) + 8^(2k + 1), then 57 also divides 7^((k + 1) + 2) + 8^(2(k + 1) + 1).
Using the assumption that 57 divides 7^(k + 2) + 8^(2k + 1), we can express:
7^(k + 2) + 8^(2k + 1) = 57a (where a is an integer)
Now, let's consider the expression for n = k + 1:
7^((k + 1) + 2) + 8^(2(k + 1) + 1) = 7^(k + 3) + 8^(2k + 3)
Using the properties of exponents, we can rewrite the expression as:
7^(k + 1) * 7^2 + 8^(2k + 1) * 8^2 = (7^2 * 7^(k + 1)) + (8^2 * 8^(2k + 1))
By factoring out the common terms, we have:
(49 * 7^(k + 1)) + (64 * 8^(2k + 1))
Now, let's substitute the expression we obtained earlier for 7^(k + 2) and 8^(2k + 1):
(49 * 57a) + (64 * 57a)
By factoring out 57 from each term, we get:
57 * (49a + 64a)
Simplifying the expression inside the parentheses:
57 * (113a)
Therefore, we have expressed 7^((k + 1) + 2) + 8^(2(k + 1) + 1) as a multiple of 57.
Step 4: Conclusion
By proving the base case and showing that the statement holds for k implies it holds for k + 1, we can conclude that the statement "57 divides 7^(n + 2) + 8^(2n + 1)" is true for all positive integers n by mathematical induction.