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Math Help - proofs by induction

  1. #1
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    proofs by induction

    2+6+10+...+(4n-2)=2(n)^2
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  2. #2
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    Show true for n=1: 4(1)-2=2(1)^2. True.

    Assume true for 2+6+10+....+(4k-2)=2k^{2}

    Show true for P_{k+1}:

    2+6+10+....+(4k-2)+(4k+2)=2k^{2}+4k+2=2(k+1)^{2}

    It is true for the P_{k+1} case.
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  3. #3
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    thanks

    Awsome
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  4. #4
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    got another

    2+4+6+....+2n = n(n+1)
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  5. #5
    Eater of Worlds
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    Quote Originally Posted by gohangoten1 View Post
    2+4+6+....+2n = n(n+1)
    Just add 2n+2 to both sides and see what that gives you on the right side.

    I bet it's n(n+1)+2n+2=(n+1)(n+2).

    Afterall, what would be the next number in the series?. 2n+2. See?
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  6. #6
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    got another

    5+10+15+...+5n= 5n(n+1)/2
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  7. #7
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    Try this one on your own. All these induction proofs with sums follow the same pattern

    Show for a base (usually n=1)
    Suppose P(n) which is "5 + 10 + 15 + ... + 5n = 5n(n+1)/2"
    Show P(n+1) supposing P(n)

    with these sums, it amounts to this for P(n+1)

    5 + 10 + 15 + ... + 5n + 5(n+1)
    but what is the bolded part equal to according to assumption? can you finish?
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  8. #8
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    i think i got it

    thanks
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