Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago

# I need help with this question. Any help is appreciate! Use mathematical induction to prove that 4n - 1 is divisible by 3 for all n >= 1?

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• 6 months ago

Did you mean 4^(n) - 1?

4^(1) - 1 = 4 - 1 = 3

4^(2) - 1 = 16 - 1 = 15

4^(3) - 1 = 64 - 1 = 63

4^(n) - 1 = 3k

4^(n) = 3k + 1

4^(n + 1) = 4 * (3k + 1)

4^(n + 1) = 12k + 4

4^(n + 1) = 12k + 3 + 1

4^(n + 1) - 1 = 12k + 3

4^(n + 1) - 1 = 3 * (4k + 1)

Well that looks familiar. Compare it to 4^(n) - 1 = 3k and assign no real values to n or k and we have the same basic form. And since 4^(n) - 1 is divisible by 3 and 4^(n + 1) - 1 is also divisible by 3, then that means that 4^(n + 2) - 1 will also be divisible by 3 and so on.

• Max6 months agoReport

The format didn't carry over to yahoo answers when i copy pasted, but this is the answer i was looking for. Thank you so much!

• 6 months ago

4(2)-1=7 prime number

4(8)-1=31prime number

etc.,