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Math Help - induction problem,

  1. #1
    Junior Member
    Joined
    Jan 2009
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    57

    induction problem,

    I have been proving a couple of induction problems.
    I am struggling when there are sequences on both sides of the equals.

    How do I prove by induction.
    1^3+2^3+3^3+ ... +n^3={(1+2+3+...+n)}^2

    demonstrating for k=1 is trivial.
    Assume for k, and then prove for (k+1).

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

    using \sum\limits_1^n n = \frac{n(n+1)}{2}

    ={[\frac{k(k+1)}{2} + (k+1)]}^2

    I have tried expanding this, it quickly get's horrible, but also doesn't seem closer to solving the problem.

    Thanks
    Regards
    Craig.
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Since 1 + 2 + ... + n = \frac{1}{2}n(n+1)
    (1 + 2 + ... + n)^2 = \bigg{(}\frac{1}{2}n(n+1)\bigg{)}^2

    So... You proved it for 1. Assume for n, prove for n+1

    1^3 + 2^3 + ... n^3 + (n+1)^3 = (1^3 + 2^3 + ... + n^3) + (n+1)^3

    = (1 + 2 + ... + n)^2 + (n+1)^3

    =\bigg{(}\frac{1}{2}n(n+1)\bigg{)}^2 + (n+1)^3
    = \frac{1}{4}(n+1)^2(n^2 + 4n + 4)
    = \bigg{(}\frac{1}{2}(n+1)(n+2)\bigg{)}^2
    = (1+2+...+(n+1))^2.

    Hence proved by induction
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