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Math Help - Mathematical induction

  1. #1
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    Mathematical induction

    Ok, so I'm trying to use mathematical induction to prove 1^3+2^3+...+n^3=[n(n+1)/2]^2 whenever n is a positive integer. I think I have it, but I'm not sure, can any of you help me to make sure I'm on the right track? Thanks!

    Basis step: P(1) states that the sum of the first one integer to the third power is [n(n+1)/2]^2. This is true since the sum of the first one integer to the third power is 1. [1(1+1)/2]^2= [1(2)/2]^2=(2/2)^2=1

    Induction Step: Assume P(n) is true for a positive integer n that is

    [2(2+1)/2]^2 [3(3+1)/2]^2
    [2(3)/2]^2 [3(4)/2]^2
    (6/2)^2 (12/2)^2
    3^2 6^2
    9=1^3+2^3 36=1^3+2^3+3^3
    9=1+8 =9 36=1+8+27=36

    Conclusion: The sum of the first n positive integers to the third power is given by [n(n+1)/2]^2 for all integers n>=1
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sjenkins View Post
    Ok, so I'm trying to use mathematical induction to prove 1^3+2^3+...+n^3=[n(n+1)/2]^2 whenever n is a positive integer. I think I have it, but I'm not sure, can any of you help me to make sure I'm on the right track? Thanks!

    Basis step: P(1) states that the sum of the first one integer to the third power is [n(n+1)/2]^2. This is true since the sum of the first one integer to the third power is 1. [1(1+1)/2]^2= [1(2)/2]^2=(2/2)^2=1

    Induction Step: Assume P(n) is true for a positive integer n that is

    [2(2+1)/2]^2 [3(3+1)/2]^2
    [2(3)/2]^2 [3(4)/2]^2
    (6/2)^2 (12/2)^2
    3^2 6^2
    9=1^3+2^3 36=1^3+2^3+3^3
    9=1+8 =9 36=1+8+27=36

    Conclusion: The sum of the first n positive integers to the third power is given by [n(n+1)/2]^2 for all integers n>=1
    um, no. you are not doing mathematical induction here

    if you assume P(n) is true (by the way, you need to state what P(n) is) then you assume:

    1^3 + 2^3 + \cdots + n^3 = \Bigg[ \frac {n(n + 1)}2 \Bigg]^2

    now you need to show P(n + 1) holds. that is, you need to show that P(n) holds implies 1^3 + 2^3 + \cdots (n + 1)^3 = \Bigg[  \frac {(n + 1)(n + 2)}2\Bigg]^2

    Hint: add (n + 1)^3 to both sides of P(n) then expand and simplify. or alternatively, you may find it easier to work the other way, start by expanding the right side of the P(n + 1) statement, and show that you can get the left side
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  3. #3
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    Unhappy

    Is there any way someone can explain this further to me? I am obviously not getting it.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sjenkins View Post
    Is there any way someone can explain this further to me? I am obviously not getting it.
    since we have verified the base case, assume P(n) is true. now we need to show P(n + 1) is true.

    since P(n) is true:

    1 + 2^3 + \cdots + n^3 = \Bigg[ \frac {n(n + 1)}2 \Bigg]^2

    add (n + 1)^3 to both sides

    \Rightarrow 1 + 2^3 + \cdots + n^3 + (n + 1)^3 = \Bigg[ \frac {n(n + 1)}2 \Bigg]^2 + (n + 1)^3

    now, simplify the right hand side to get \Bigg[ \frac {(n + 1)(n + 2)}2 \Bigg]^2

    that would prove: 1 + 2^3 + \cdots + n^3 + (n + 1)^3 = \Bigg[ \frac {(n + 1)(n + 2)}2 \Bigg]^2

    which is exactly the statement P(n + 1)
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  5. #5
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    Thank you so much for the help, now I can see!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sjenkins View Post
    Thank you so much for the help, now I can see!
    good
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