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Math Help - Prove

  1. #1
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    Prove

    1.
    1^3 + 2^3 + ... + n3 = (n(n+1)/2 )^2 for any positive integer n.

    2.
    3^n < n! for all n > 6 with n ∈ N.

    mathematical induction to prove these?
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  2. #2
    o_O
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    Yes that's the way to go. Without going through all the formalities ...

    #1: Assume it holds for k. We want to show that it holds for k+1 as well, that is, we want to show that 1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2
    So for your inductive step: {\color{red}1^3 + 2^3 + \cdots + k^3} + (k+1)^3  = {\color{red}\left(\frac{k(k+1)}{2}\right)^2} + (k+1)^3 = \cdots

    #2: Again, assume it holds for k for all k > 6, so we're assuming that 3^k < k!. We then want to show that it is true for k+1 as well, that is, 3^{k+1} < (k+1)!

    Note that: 3^{k+1} = 3^k{\color{blue}3^1} < 3^k{\color{blue}(k+1)} < \cdots

    We can assume the blue because we're given that the statement is true for values strictly greater than 6. So obviously, 3 < 6 < k+1. Hopefully you'll see the conclusion directly after using the inductive step.
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