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Thread: (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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    $\displaystyle (k+1)!=(k+1) \cdot k \cdot (k-1) \cdot ... \cdot 2 \cdot 1$

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

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

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

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

    Taking $\displaystyle n=k$, we have

    $\displaystyle k!>2^k$

    Multiplying by k+1, we have

    $\displaystyle k!(k+1)>2^k(k+1)$ or $\displaystyle (k+1)!>2^k(k+1)$

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

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

    By transitivity, we have

    $\displaystyle (k+1)!>2^{k+1}$

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