Results 1 to 3 of 3

Thread: quick question about inductive proof

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    129

    quick question about inductive proof

    Prove that: $\displaystyle 3^n < n!$ n is an integer greater than 6.

    Basic step:
    $\displaystyle P(7): 3^7 < 7! $
    2187 < 5040

    Inductive Steps:
    hypothesis: $\displaystyle P(k): 3^k < k!$
    conclusion: $\displaystyle P(k+1): 3^{k+1} < (k+1)!$

    Proof: $\displaystyle 3^(k+1) = 3.3^k < 3.k! < (k+1)k! = (k+1)!$


    why is this one: $\displaystyle (k+1)k!$ equal to $\displaystyle (k+1)!$?

    Is there any better way to prove it?
    Last edited by zpwnchen; Oct 17th 2009 at 08:28 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    If you have more than one character in an exponent, set off the whole exponent in braces.
    [tex]3^{(k+1)}[/tex] gives $\displaystyle 3^{(k+1)}$ instead of $\displaystyle 3^(k+1)$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by zpwnchen View Post
    Prove that: $\displaystyle 3^n < n!$ n is an integer greater than 6.

    Basic step:
    $\displaystyle P(7): 3^7 < 7! $
    2187 < 5040

    Inductive Steps:
    hypothesis: $\displaystyle P(k): 3^k < k!$
    conclusion: $\displaystyle P(k+1): 3^(k+1) < (k+1)!$

    Proof: $\displaystyle 3^(k+1) = 3.3^k < 3.k! < (k+1)k! = (k+1)!$


    why is this one: $\displaystyle (k+1)k!$ equal to $\displaystyle (k+1)!$?

    Is there any better way to prove it?
    $\displaystyle k! = 1\cdot 2 \cdot 3 \cdot ... \cdot k$
    $\displaystyle (k+1)! = 1\cdot 2 \cdot 3 \cdot ... \cdot k \cdot (k+1)$
    $\displaystyle (k+1)k! = (k+1)(1\cdot 2 \cdot 3 \cdot ... \cdot k) = 1\cdot 2 \cdot 3 \cdot ... \cdot k \cdot (k+1) = (k+1)!$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Dec 29th 2010, 07:38 AM
  2. Inductive proof
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: Dec 8th 2009, 01:11 AM
  3. Ugh, another inductive proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Nov 13th 2008, 02:56 PM
  4. Inductive proof
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: Apr 28th 2008, 06:24 PM
  5. inductive proof question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Sep 15th 2007, 05:22 AM

Search Tags


/mathhelpforum @mathhelpforum