Question

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

Answer 1

Below.

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.