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Math Help - Mathematical Induction Proofs

  1. #1
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    Unhappy Mathematical Induction Proofs

    I have absolutely no idea how to do this problem. PLEASE SHOW ME HOW TO DO IT!!!!!

    Verify by mathematical induction.

    (1^3)+(3^3)+(5^3)+...+(2n-1)^3 = n^2[2(n^2)-1]
    Last edited by zuic; April 6th 2008 at 07:17 AM.
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  2. #2
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    Quote Originally Posted by zuic View Post
    I have absolutely no idea how to do this problem. PLEASE HELP!!!!!

    Verify by mathematical induction.

    (1^3)+(3^3)+(5^3)+...+(2n-1)^3 = n^2[2(n^2)-1]
    Most mathematical induction problems are done the same way. Some may require a little ingenuity on the second step than others, but the overall format is still the same.


    start by verifying that P(1) is true.

    then assume P(n) is true.

    show that P(n) implies P(n + 1) is true.

    Then the conclusion will follow by Mathematical induction.

    care to take a shot at it?
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  3. #3
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    Hello,

    Firstly, prove that it's true for n=0, then, prove that : (1^3)+(3^3)+(5^3)+...+(2n-1)^3+(2(n+1)-1)^3=(n+1)^2[2(n+1)^2-1], supposing that (1^3)+(3^3)+(5^3)+...+(2n-1)^3 = n^2[2(n^2)-1] (induction hypothesis)

    This is a proof by induction.

    So, you have (1^3)+(3^3)+(5^3)+...+(2n-1)^3+(2(n+1)-1)^3

    =\underbrace{(1^3)+(3^3)+(5^3)+...+(2n-1)^3}_{n^2[2(n^2)-1]} +(2n+1)^3

    =n^2[2n^2-1]+(2n+1)^3

    Develop and retrieve (n+1)^2(2(n+1)^2-1)
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  4. #4
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    First, test your formula for a trivial value of n

    [Math] \sum_{i=1}^{1}{(2i-1)^3} = 1 = 1^2(2 \times 1^2-1) [/tex]

    Assume it work for all integer value of n up to a a particular value say k.

    so we assume that [Math] \sum_{i=1}^{k}{(2i-1)^3} = k^2(2 k^2-1) [/tex] is true.

    then attempt to prove that the formula is also true for k+1.

    [Math] \sum_{i=1}^{k+1}{(2i-1)^3} = \sum_{i=1}^{k}{(2i-1)^3} + (2(k+1) -1)^3 [/tex]

    [Math] \Rightarrow k^2(2 k^2-1) +(2k +1)^3 [/tex]

    ..... (omitting some algebra) .....

    [Math] 2k^4 + 8k^3 + 11k^2 + 6k +1 [/tex]

    now when you factor this remember your expecting k+1 to be a factor, (it's not cheating to use your expected result to help factor an expression).

    [Math]\Rightarrow (k+1)(2k^3+6k^2+5k+1) [/tex]
    [Math]\Rightarrow (k+1)^2(2k^2+4k+1) [/tex]
    [Math]\Rightarrow (k+1)^2(2(k+1)^2 -1 ) [/tex]

    If true of k , it is also true for k+1, we have shown that it is true for k=1 so it is therefore. also true for k=2 and hence all integers.

    Bobak
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  5. #5
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    1^{3}+3^{3}+5^{3}+.....+(2n-1)^{3}=n^{2}(2n^{2}-1)

    Check for n=1: 1^{3}=(1)^{2}(2(1)^{2}-1), TRUE.

    Assume P_{k} is true, then the induction hypothesis is:

    1^{3}+3^{3}+5^{3}+....+(2k-1)^{3}=k^{2}(2k^{2}-1)

    Now we have to show that P_{k+1} is true. Add (2k+1)^{3} to both sides, which is the next odd cube.

    1^{3}+3^{3}+5^{3}+.....+(2k-1)^{3}+(2k+1)^{3}=k^{2}(2k^{2}-1)+(2k+1)^{3}

    k^{2}(2k^{2}-1)+(2k+1)^{3}=2k^{4}+8k^{3}+11k^{2}+6k+1 =(k+1)^{2}(2k^{2}+4k+1)=(k+1)^{2}(2(k+1)^{2}-1)

    and the induction is complete.
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