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Math Help - (k+1)!

  1. #1
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    (k+1)!

    How is this evaluated. Is it k(k+1) or is it (k+1)(k+2). This is the problem if anyone wants to try it out.

    Prove by induction on n that n!>2^n for all integers n such that n >or= 4
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  2. #2
    Senior Member DivideBy0's Avatar
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    (k+1)!=(k+1) \cdot k \cdot (k-1) \cdot ... \cdot 2 \cdot 1

    Proposition: n!>2^n for n\geq 4.

    Proof: We will prove the proposition by induction on a variable n.

    If n=4, we have 4!>2^4 or 24>16, which is true.

    Assume: n! >2^n for 4\leq n \leq k

    Taking n=k, we have

    k!>2^k

    Multiplying by k+1, we have

    k!(k+1)>2^k(k+1) or (k+1)!>2^k(k+1)

    And since k \geq 4, the minimum value we can have for k+1 is 5, so we have

    2^k(k+1)>2^k \cdot 2 or 2^k(k+1)>2^{k+1}

    By transitivity, we have

    (k+1)!>2^{k+1}

    Hence, (k+1)!>2^{k+1} for n \geq 4, QED
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