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Math Help - Help with Proofs by Induction

  1. #1
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    Help with Proofs by Induction

    Hello everyone. I am really struggling to wrap my head around proofs by induction. I have a problem that I need help understanding.

    1. Prove: 4 + 8 + 12 +...+ 4n = 2n2 + 2n for all integers n >= 1

    Does the base case always have to equal 1? Or is the base case 4. After doing some algebra I ended up with 4.

    After the I get the base case I'm having trouble understanding how to come up with the inductive hypothesis and how to put it all in a proof.
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  2. #2
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    Re: Help with Proofs by Induction

    Quote Originally Posted by troe View Post
    After the I get the base case I'm having trouble understanding how to come up with the inductive hypothesis and how to put it all in a proof.
    I recommend studying this description of a proof by induction.
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  3. #3
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    Re: Help with Proofs by Induction

    Quote Originally Posted by troe View Post
    Hello everyone. I am really struggling to wrap my head around proofs by induction. I have a problem that I need help understanding.

    1. Prove: 4 + 8 + 12 +...+ 4n = 2n2 + 2n for all integers n >= 1

    Does the base case always have to equal 1? Or is the base case 4. After doing some algebra I ended up with 4.
    You ended up with 4 for what? The problem says "for all integers n>= 1". You are doing induction on n.
    when n= 1, you have 4= 2(1)2+ 2(1) which is true.

    Assume that, for some k, 4+ 8+ 12+ ...+ 4k= 2k2+ 2k. Then the k+1 case is 4+ 8+ 12+ ...+ 4k+ 4(k+1)= 2k+ 2k+ 4k+ 4
    Can you show that that is equal 2(k+1)2+ 2(k+1)?

    After the I get the base case I'm having trouble understanding how to come up with the inductive hypothesis and how to put it all in a proof.
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