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Math Help - Proof by Induction

  1. #1
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    Proof by Induction

    Hey all,
    can anyone show me how I can prove a statement like this by Induction. Thanks.

    1^3+2^3+3^3+...+n^3=\frac{1}{4}n^2(n+1)^2 for all  n \in \mathbb{Z}
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  2. #2
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    Well, your n is natural unzero. (the sum starts with 1)
    Read this: Mathematical induction - Wikipedia, the free encyclopedia

    First step - verify if for n=1 that equality is true.
    Second step (Induction) - we (you) suppose that k verify the equality and prove that then k+1 verify too.
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  3. #3
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    Hey,

    Stuck on the Induction step where I must rewrite the left-hand side..
    ..(1^3+2^3+3^3+...+n^3)+(n^3+1)=1/4*(n^3+1)(n^3+2)^2

    EDIT: Should be (1^3+2^3+3^3+...+n^3)+(n+1)=1/4*(n+1)^2((n+1)+1)^2 and is straight forward to prove from there. Thanks for the help..
    Last edited by Oiler; April 23rd 2011 at 10:28 PM.
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  4. #4
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    Quote Originally Posted by Oiler View Post
    Hey,

    Stuck on the Induction step where I must rewrite the left-hand side..
    ..(1^3+2^3+3^3+...+n^3)+(n^3+1)=1/4*(n^3+1)(n^3+2)^2

    EDIT: Should be (1^3+2^3+3^3+...+n^3)+(n+1)=1/4*(n+1)^2((n+1)+1)^2 and is straight forward to prove from there. Thanks for the help..
    No,

    you need

    P(k)

    1^3+2^3+3^3+.....+k^3=(1/4)k^2(k+1)^2

    P(k+1)

    1^3+2^3+3^3+....+k^3+(k+1)^3=(1/4)(k+1)^2(k+2)^2

    and so you will be trying to prove that

    (1/4)k^2(k+1)^2+(k+1)^3=(1/4)(k+1)^2(k+2)^2

    where both sides have a common factor of (k+1)^2
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  5. #5
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    Induction is not necessary...

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  6. #6
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    Unfortunately Prove It, as cool as it is,
    you get zero marks for answering the exam question
    "Prove using induction that......."
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    No,

    you need

    P(k)

    1^3+2^3+3^3+.....+k^3=(1/4)k^2(k+1)^2

    P(k+1)

    1^3+2^3+3^3+....+k^3+(k+1)^3=(1/4)(k+1)^2(k+2)^2

    and so you will be trying to prove that

    (1/4)k^2(k+1)^2+(k+1)^3=(1/4)(k+1)^2(k+2)^2

    where both sides have a common factor of (k+1)^2
    Since there's a cube on the LHS reducing it to match the RHS suddenly doesn't seem too trivial..

    I get stuck showing the LHS = RHS

    ..(n^k(k+1)^2+4(k+1)^3))/4..

    Thanks for the help..
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  8. #8
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    Write the LHS as

    (1/4)k^2(k+1)^2+(k+1)(k+1)^2

    Then divide both sides by (k+1)^2 to get

    (1/4)k^2+k+1 = (1/4)(k+2)^2 =(1/4)(k^2+4k+4)

    QED.
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