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Math Help - Mathematical Induction help

  1. #1
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    Mathematical Induction help

    1^3+2^3+...+n^3 = [n(n+1)/2]^2 n>= 1

    If anyone has a link to write in mathematical nottaion on this forum, i will appreciate it, thanks.
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  2. #2
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    Hi

    Have a look to the LATEX tutorial in the LATEX forum

    Let P(n) be 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2

    Check that P(1) is true

    Suppose that P(k) is true and show that P(k+1) is true
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  3. #3
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    Thanks for the help, but i need help on the induction stage, when i do n+1. Here is what i have.

     \frac{n(n+1)^2+2(n+1)^3}{2} = \frac{((n+1(n+2)))}{2}^2
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  4. #4
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    Quote Originally Posted by tokio View Post
    Thanks for the help, but i need help on the induction stage, when i do n+1. Here is what i have.

     \frac{n(n+1)^2+2(n+1)^3}{2} = \frac{((n+1(n+2)))}{2}^2
    You suppose that P(n) is true, which means that 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2

    You want to prove that P(n+1) is true : 1^3+2^3+...+n^3+(n+1)^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2

    1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+(n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3

    Factor (n+1)/4
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  5. #5
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    Quote Originally Posted by running-gag View Post
    You suppose that P(n) is true, which means that 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2

    You want to prove that P(n+1) is true : 1^3+2^3+...+n^3+(n+1)^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2

    1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+(n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3

    Factor (n+1)/4

    Im still stuck on the induction stage, It seems like if i still factor that i still cant get both sides of the quation to be logically equilvalent.
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  6. #6
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    Proof by induction

    Hello tokio
    Quote Originally Posted by tokio View Post
    Im still stuck on the induction stage, It seems like if i still factor that i still cant get both sides of the quation to be logically equilvalent.
    Here's the bit you need:

    \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3

    = \frac{n^2\color{red}(n+1)^2}{\color{red}4} + \frac{4(n+1)\color{red}(n+1)^2}{\color{red}4}

    = \color{red}\frac{(n+1)^2}{4}\color{black}(n^2 + 4(n+1))

    = \frac{(n+1)^2}{4}(n+2)^2

    Can you complete it now?

    Grandad
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  7. #7
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    Quote Originally Posted by Grandad View Post
    Hello tokioHere's the bit you need:

    \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3

    = \frac{n^2\color{red}(n+1)^2}{\color{red}4} + \frac{4(n+1)\color{red}(n+1)^2}{\color{red}4}

    = \color{red}\frac{(n+1)^2}{4}\color{black}(n^2 + 4(n+1))

    = \frac{(n+1)^2}{4}(n+2)^2

    Can you complete it now?

    Grandad
    Thanks for the help but i still cant believe im confused. Im assume that your just working onthe right hand side, i dont see where you proved for (n+1).
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  8. #8
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    Proof by induction

    Hello tokio
    Quote Originally Posted by tokio View Post
    Thanks for the help but i still cant believe im confused. Im assume that your just working onthe right hand side, i dont see where you proved for (n+1).
    That's correct. I thought it was just the manipulation of the RHS that you couldn't do. The complete proof is:

    Let P(n) be the propositional function: 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2

    Then P(n) \Rightarrow 1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+ (n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3

    = \dots (as in my previous posting)

    = \frac{(n+1)^2}{4}(n+2)^2

    = \left(\frac{(n+1)([n+1]+1)}{2}\right)^2

    \Rightarrow P(n+1)

    P(1) is 1^3 = \left(\frac{1 \cdot 2}{2} \right)^2, which is true. Hence P(n),  \forall n \in \mathbb{N}.

    Grandad

    PS If you want to see the LaTeX that's been used to generate any of the expressions, just click on them and a little window will open showing you the code.
    Last edited by Grandad; March 6th 2009 at 10:13 PM. Reason: Add PS
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