prove by mathematical induction that 7^n+4^n+1 is divisible by 6
7^n + 4^n + 1
1. test for n = 1
7^1 + 4^1 + 1 = 12 , which is divisible by 6
2. assume it is true for n = k , that is, assume that
7^k + 4^k + 1 is divisible by 6
3. then show that 7^(k+1) + 4^(k+1) - 1 is divisible by 6
use the number property that if both A and B are divisble by c
then A-B is divisible by c
e.g. 156 and 117 are both divisible by 13
then is 156-117 or 39 divisible by 13 ? YES
so ...
7^(k+1) + 4^(k+1) + 1 - (7^k + 4^k + 1)
= 7^(k+1) + 4^(k+1) + 1 -7^k - 4^k - 1
= 7^k(7-1) + 4^k(4-1)
= 6(7^k) + 3(4^k)
clearl 6(7^k) is a multiple of 6 , thus divisible by 6
and in 3(4^k) , the 4^k must be even and any even times 3 is divisible by 6
so we have shown that the result is divisible by 6
(the sum of multiples of 6 must be divisible by 6)
To prove that 7^n + 4^n + 1 is divisible by 6 for all positive integers n using mathematical induction, we need to follow these steps:
Step 1: Base Case
Initially, we need to show that the statement holds true for the smallest possible value of n, which is n = 1.
For n = 1, we have:
7^1 + 4^1 + 1 = 7 + 4 + 1 = 12
Since 12 is divisible by 6, the statement holds true for the base case.
Step 2: Inductive Hypothesis
Assume that the statement holds true for some arbitrary positive integer k. That is:
7^k + 4^k + 1 is divisible by 6.
Step 3: Inductive Step
Now, we need to prove that if the statement holds for k, it also holds for k+1.
We need to show that 7^(k+1) + 4^(k+1) + 1 is divisible by 6.
First, let's write the expression for k+1:
7^(k+1) + 4^(k+1) + 1 = 7^k * 7^1 + 4^k * 4^1 + 1
Using the assumption from the inductive hypothesis, we know that 7^k + 4^k + 1 is divisible by 6. Therefore, we can write:
7^k + 4^k + 1 = 6m (where m is some positive integer)
Now, substitute this expression into the equation for k+1:
7^(k+1) + 4^(k+1) + 1 = (6m - 1) * 7 + (6m - 1) * 4 + 1
Simplify the expression further:
= 42m - 7 + 24m - 4 + 1
= 66m - 10
Since 66m - 10 can be factored as 6(11m - 1), which is divisible by 6, we can conclude that 7^(k+1) + 4^(k+1) + 1 is divisible by 6.
Therefore, by the principle of mathematical induction, 7^n + 4^n + 1 is divisible by 6 for all positive integers n.