prove by mathematical induction that 7n+4n+1 is divisible by 6
As written, it's obviously false. Try n=2
Did you mean 7n^2+4n+1? Nope; false for n=2
Got some error here
its 7^n+4^n+1
Steve
Ah; that's a lot nicer.
it's true for n=1.
So, assume it's true for n=k.
7^(k+1) + 4^(k+1) + 1
= 7*7^k + 4*4^k + 1
= (1+6)*7^k + (1+3)*4^k + 1
= (7^k+4^k+1) + 6*7^k + 3*4^k
obviously, 6*7^k is divisible by 6
4^k is even, so 3*4^k is divisible by 6
So, since we're adding three items which are all multiples of 6, the whole is a multiple of 6.
Thus, the induction step holds.
To prove that the expression 7n + 4n + 1 is divisible by 6 for all positive integers n, we will use mathematical induction.
Step 1: Base Case
We start by checking the base case, which is usually n = 1. Let's substitute n = 1 into the expression:
7(1) + 4(1) + 1 = 7 + 4 + 1 = 12.
Since 12 is divisible by 6, our base case is satisfied.
Step 2: Inductive Hypothesis
Assume that for some positive integer k, the expression 7k + 4k + 1 is divisible by 6. This is our inductive hypothesis.
Step 3: Inductive Step
We need to prove that if our hypothesis is true for k = m (where m is any positive integer), then it must also be true for k = m + 1.
Substituting k = m + 1 into the expression:
7(m+1) + 4(m+1) + 1 = 7m + 7 + 4m + 4 + 1 = (7m + 4m) + (7 + 4 + 1) = 11m + 12.
Now, we want to show that 11m + 12 is divisible by 6.
Step 4: Prove the Inductive Step
To prove that 11m + 12 is divisible by 6, we need to show that there exists an integer p such that (11m + 12) = 6p.
We can rewrite 11m + 12 as 6(m + 2) + (m - 1).
The first part, 6(m + 2), is clearly divisible by 6 since it is a multiple of 6.
Now, let's focus on the second part, (m - 1).
Since our inductive hypothesis assumes that 7k + 4k + 1 is divisible by 6 for any positive integer k, and k = m, we know that 7m + 4m + 1 is divisible by 6.
Therefore, the quantity (7m + 4m + 1) is divisible by 6, and multiplying it by 11 will still be divisible by 6.
So, we can write (7m + 4m + 1) as 6p for some integer p.
Therefore, (11m + 12) = 6(m + 2) + (m - 1) = 6(m + 2) + 6p, which is clearly divisible by 6.
Since the statement holds for k = m + 1, based on our inductive hypothesis, we have proven that for all positive integers n, 7n + 4n + 1 is divisible by 6.