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Proof by Induction Problem

  1. Oct 23, 2009 #1
    Nevermind, got the answer :D

    1. The problem statement, all variables and given/known data
    I need to prove using mathematical induction that:

    [tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]


    2. Relevant equations

    [tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]


    3. The attempt at a solution

    Ok, I have proved that the statement is true when n=1 and have gone on to assume that it must be true for any arbitrary value of n. I chose n=p. Thus:

    [tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p ^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}[/tex]

    Where i am stuck is that I know i need to show that if the statement is true when n=p then it must be true for n=p+1. I have gotten as far as:

    [tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}[/tex]
     
    Last edited: Oct 23, 2009
  2. jcsd
  3. Oct 23, 2009 #2

    whs

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    [tex]
    \sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}
    [/tex]

    Factor out a (p+1)^2, and see what you see.
     
  4. Oct 23, 2009 #3
    Re: Nevermind, got the answer :D

    have you tried just expanding it out and seeing what you get?
    the induction step is simple, you got that - the rest is just algebra
     
  5. Oct 23, 2009 #4
    Basically, you should expand, collect terms, and reduce. You should be able to do it that way.
     
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