using mathematical induction prove the following?
6n-1isdivisible by 5,for n>0 ?
Clearly, 6*2 - 1 = 12-1 = 11
is not divisible by 5.
So, I assume you meant 6^n - 1 is divisible by 5.
Check P(1):
6^1 - 1 = 6-1 = 5, is divisible by 5. So, P(1) is true.
Assume P(k). Then
6^(k+1) - 1 = 6*6^k - 1
= 5*6^k + 6^k - 1
Now, we know that 6^k - 1 is a multiple of 5, say 5m
So, 5*6^k + 5m = 5(6^k+m) is divisible by 5.
So, we have P(1) and P(k) ==> P(k+1)
so, P(k) is true for all k > 0.
To prove that 6n-1 is divisible by 5 for n > 0 using mathematical induction, we need to follow these steps:
Step 1: Show that the statement is true for the base case.
- In this case, the base case would be n = 1.
- Substitute n = 1 into the expression and check if the result is divisible by 5.
6(1) - 1 = 6 - 1 = 5
- Since 5 is divisible by 5, the statement is true for the base case.
Step 2: Assume the statement is true for some arbitrary positive integer, k.
- Assume the statement is true for n = k: 6k - 1 is divisible by 5.
Step 3: Prove that the statement is also true for the next positive integer, k + 1.
- Substitute n = k + 1 into the expression and simplify:
6(k + 1) - 1 = 6k + 6 - 1 = (6k - 1) + 5
- We know that 6k - 1 is divisible by 5 from the assumption in Step 2.
Let's say 6k - 1 = 5m for some positive integer m.
- Substituting the value of 6k - 1 in the expression, we get:
(6k - 1) + 5m = 5m + 5 = 5(m + 1)
- Since m + 1 is a positive integer, we can see that 5(m + 1) is divisible by 5.
Conclusion: By proving that the statement is true for the base case (n = 1) and showing that if it is true for k, then it is also true for k+1, we have established the proof by mathematical induction. Therefore, 6n - 1 is divisible by 5 for n > 0.