Using Mathematical Induction, prove the following:
a) 6^n - 1 is divisible by 5, for n>_0.
To prove that 6^n - 1 is divisible by 5 for n ≥ 0 using mathematical induction, we will follow the following steps:
Step 1: Base Case
For n = 0, 6^0 - 1 = 1 - 1 = 0. Since 0 is divisible by 5, the statement holds true for the base case.
Step 2: Inductive Hypothesis
Assume that the statement is true for some arbitrary positive integer k, i.e., assume that 6^k - 1 is divisible by 5.
Step 3: Inductive Step
We will prove that the statement is also true for k + 1, i.e., we will prove that 6^(k + 1) - 1 is divisible by 5 using the assumption made in the inductive hypothesis.
Starting with 6^(k + 1) - 1:
= 6 * 6^k - 1
= 6 * 6^k - 6 + 5
= 6 * (6^k - 1) + 5
Since we assumed that 6^k - 1 is divisible by 5, let's substitute it in the above expression:
= 6 * (5m) + 5 [where m is an integer, because 6^k - 1 is divisible by 5]
Now, let's simplify:
= 30m + 5
= 5(6m + 1)
Since 6m + 1 is an integer, we have expressed 6^(k + 1) - 1 in terms of 5 multiplied by an integer. Hence, 6^(k + 1) - 1 is divisible by 5.
Step 4: Conclusion
Since the statement is true for the base case (n = 0) and we have proven that if it is true for k, then it is also true for k + 1, we can conclude that 6^n - 1 is divisible by 5 for n ≥ 0 using mathematical induction.
To prove that 6^n - 1 is divisible by 5 for n≥0 using mathematical induction, we need to follow these steps:
Step 1: Base Case
First, we need to verify the statement for the base case, which is n = 0.
When n = 0, we have 6^0 - 1 = 1 - 1 = 0. Since 0 divided by 5 is equal to 0 with no remainder, the base case is valid.
Step 2: Inductive Hypothesis
Assume that the statement is true for some value k, where k≥0.
So, we assume that 6^k - 1 is divisible by 5.
Step 3: Inductive Step
Now, we need to prove the statement is also true for the next value, which is k+1.
Consider 6^(k+1) - 1, which can be written as 6^k * 6 - 1.
By the inductive hypothesis, we know that 6^k - 1 is divisible by 5.
Let's represent 6^k - 1 as 5m, where m is a positive integer.
Now, 6^(k+1) - 1 = 6^k * 6 - 1 = 6 * 5m - 1 = 30m - 1.
Since 30m is divisible by 5, the remainder, when subtracting 1, is also divisible by 5.
Therefore, 6^(k+1) - 1 is divisible by 5.
Step 4: Conclusion
By mathematical induction, we have shown that if the statement is true for the base case (n = 0) and if it is true for k, then it is also true for k+1. Hence, the statement is proven to be true for all n≥0.
check P(0).
6^0-1 = 0, which is divisible by 5
Assume P(k). Now,
6^(n+1)-1 = (5+1)^(n+1) - 1
Now use the Binomial Theorem to expand that.
= 5^(n+1) + C(n,1)*5^n*1 + ... + C(n,n-1)*5*1^n + 1^(n+1) - 1
= 5(5^n + C(m,1)*5^(n-1) + ... + C(n,n-1))
which is divisible by 5.