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Math Help - Inductive reasoning

  1. #1
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    Inductive reasoning

    Sort of lost on how to continue...
    Prove that the following formula is true for all n using inductive reasoning 3 + 7 + 11 + .... +(4n - 1) = n(2n + 1)

    n = 1 is true
    assume k1 is true? (4k - 1) = k(2k + 1)
    now to prove k + 1

    [4(k+1) - 1] + (4k - 1) = k(2k + 1) + [4(k+1) - 1]

    Would someone mind showing me where to go from here?
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  2. #2
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    Ok

    hi-

    i think u are on the right track.
    continue.

    Just a step and u are there!
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Sxon View Post
    Sort of lost on how to continue...
    Prove that the following formula is true for all n using inductive reasoning 3 + 7 + 11 + .... +(4n - 1) = n(2n + 1)

    n = 1 is true
    assume k1 is true? (4k - 1) = k(2k + 1)
    now to prove k + 1

    [4(k+1) - 1] + (4k - 1) = k(2k + 1) + [4(k+1) - 1]

    Would someone mind showing me where to go from here?
    you mean

    \sum_{i=1}^{n} (4i-1) = n(2n+1)

    let P(k):\sum_{i=1}^{k} (4i-1) = k(2k+1)

    P(1) is true and P(2) is true suppose that P(k) is true lats try

    P(k+1):\sum_{i=1}^{k+1} (4i-1) ={\color{blue} (k+1)(2(k+1)+1)} true or not

    \sum_{i=1}^{k+1}(4i-1)={\color{red}\sum_{i=1}^{k}(4i-1)}+(4(k+1)-1)

    and the red term equal k(2k+1) from P(k) so you will have

    \sum_{i=1}^{k+1}(4i-1)=k(2k+1)+(4(k+1)-1)=2k^2+k+(4k+3)=2k^2+5k+3=(k+1)(2k+3) and this equal the blue term the proof ends
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  4. #4
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    By the way, "proof by induction", which is what is meant here, and "inductive reasoning" are not at all the same thing!
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  5. #5
    Junior Member woof's Avatar
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    Quote Originally Posted by Sxon View Post
    Sort of lost on how to continue...
    Prove that the following formula is true for all n using inductive reasoning 3 + 7 + 11 + .... +(4n - 1) = n(2n + 1)

    n = 1 is true
    assume k is true: 3 + 7 + 11 + .... +(4k - 1) = k(2k + 1)
    now to prove k + 1

    3 + 7 + 11 + .... +(4k - 1)+[4(k+1) - 1] + (4k - 1) = k(2k + 1) + [4(k+1) - 1]

    Would someone mind showing me where to go from here?
    I've added some parts that should be there -- I can't tell if you omitted them for less typing, or weren't sure that they should be there.

    You want to show that the last thing you typed: k(2k + 1) + [4(k+1) - 1] is equal to (k+1)(2(k+1) + 1), for the inductive statement to be true.

    Show that as Amer did:
    k(2k+1)+(4(k+1)-1)=2k^2+k+(4k+3)=2k^2+5k+3=(k+1)(2k+3)
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