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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    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.,

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