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-   -   Math Induction (https://www.askmehelpdesk.com/showthread.php?t=107240)

  • Jul 6, 2007, 12:18 AM
    Terrytw
    Math Induction
    Here is what I have done so far. Can anyone offer suggestions?

    1^3 + 2^3 + 3^3 + * * * + n^3 = 1/4(n^2)[(n+1)^2]

    Let P(1) = 1 then 1^3 = 1/4(1^2)[(1+1)^2]=1 so P(1) is true.

    So we assume since it is true for P(1) then it must be true for P(k) as well and
    then need to prove true for P(k+1)

    so P(k) = 1^3 + 2^3 + 3^3 + * * * + k^3 = 1/4(k^2)[(k+1)^2] = 1/4[k(k+1)]^2

    P(k+1) = 1^3 + 2^3 + 3^3 + * * * + k^3 + (k^3+1)= 1/4(k^2)[(k+1)^2] = 1/4[(k+1)(k+1)+1]^2

    and this is where I get stuck.
  • Jul 6, 2007, 03:52 AM
    ebaines
    You want to show that P(k+1) = P(k) + (k+1)^3. You wrote (k^3+1), and that's not correct.

    If you substitute (k+1) in place of k in the formula P(k) = 1/4[k^2(k+1)^2], you can expand it out, and ultimately show that:

    P(k+1) = P(k) + (k+1)^3

    Write back if you're still having difficulty.

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