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Math Help - Prove by induction

  1. #1
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    Prove by induction

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

    mathematical induction to prove these?
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  2. #2
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    Hello, captainjapan!

    Prove by induction: . 3^n \:<\: n! \;\text{ for all }n > 6,\;\;n \in N
    Verify the base case . . .
    . . S(7)\!:\;\;3^7 \:<\:7! \quad\Rightarrow\quad 2187 \:<\:5040 . . . True


    Assume S(k)\!:\;\;3^k \:<\:k!


    Multiply both sides by 3: . 3^k\cdot3 \:<\:3\cdot k! \quad\Rightarrow\quad 3^{k+1} \:<\:3\cdot k! .[1]

    . . Since k > 6\!:\;\;3 \:<\:k

    . . Multiply by k!\!:\;\;3\cdot k! \:<\:k\cdot k! \quad\Rightarrow\quad 3\cdot k! \:<\:(k+1)! .[2]


    Combine [1] and [2]: . 3^{k+1} \:<\:3\cdot k! \:<\:(k+1)!

    . . Therefore: . 3^{k+1}\:<\:(k+1)!


    And we have proved S(k+1) . . . The inductive proof is complete.

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