Question

Part II Prove that 7" – 1 is a multiple of 6 for all n EN (Use Mathematical Induction!)

Answer 1

Answer:

Below.

Step-by-step explanation:

To prove this for mathematical induction, we will need to prove:

Part 1)  That $7^n-1$ is a multiple of 6 for n=1.

Part 2) That, if by assuming $7^{n}-1$ is a multiple of 6, then showing $7^{n+1}-1$ is a multiple of 6.

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Part 1) If n=1, we have $7^n-1=7^1-1=7-1=6$ where 6 is a multiple of 6 since 6 times 1 is 6.

Part 2) A multiple of 6 is the product of 6 and k where k is an integer.  So let's assume that there is a value k such that $7^n-1=6k$ for some number natural number $n$.

We want to show that $7^{n+1}-1$ is a multiple of 6.

$7^{n+1}-1$

$7^n7^1-1$

$(7)7^n-1$

$(7)7^{n}-7+6$

$7(7^{n}-1)+6$

$7(6k)+6$ (this is where I applied my assumption)

$6[7k+1]$ (factoring with the distributive property)

Since 7k+1 is an integer then 6(7k+1) means that $7^{n+1}-1$ is a multiple of 6.

This proves that $7^n-1$ is a multiple of 6 for all natural n.