Results 1 to 3 of 3

Math Help - induction - I got most of the proof, just cant finish

  1. #1
    Member pberardi's Avatar
    Joined
    Dec 2008
    Posts
    85

    induction - I got most of the proof, just cant finish

    Prove that for every natural number greater than or equal to 6, (n+1)^2 is less than or equal to 2^n

    Let S = {n in N s.t. n>=6, (n+1)^2 <= 2^n }

    Show 6 is in S 7^2 <= 2^7
    Assume n is in S
    Show n+1 is in S
    (n+2)^2 <= 2^n +1
    n^2 + 4n + 4 <= 2^n +1 HELP!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello pberardi
    Quote Originally Posted by pberardi View Post
    Prove that for every natural number greater than or equal to 6, (n+1)^2 is less than or equal to 2^n

    Let S = {n in N s.t. n>=6, (n+1)^2 <= 2^n }

    Show 6 is in S 7^2 <= 2^7
    Assume n is in S
    Show n+1 is in S
    (n+2)^2 <= 2^n +1
    n^2 + 4n + 4 <= 2^n +1 HELP!
    n \in S \Rightarrow (n+1)^2 \le 2^n

    \Rightarrow n^2 + 2n+1 \le 2^n

    \Rightarrow n^2 + 4n+4 \le 2^n + 2n+3

    \Rightarrow (n+2)^2 \le 2^n + 2+2n+1

    \Rightarrow (n+2)^2 \le 2^n + n^2 + 2n + 1, since 2 < n^2 for n \ge 6

    \Rightarrow (n+2)^2 \le 2^n + (n+1)^2

    \Rightarrow (n+2)^2 \le 2^n +2^n, since (n+1)^2 \le 2^n

    \Rightarrow (n+2)^2 \le 2.2^n = 2^{n+1}

    Incidentally, to show 6 \in S, you should write 7^2 \le 2^6, not 2^7

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member pberardi's Avatar
    Joined
    Dec 2008
    Posts
    85
    Thanks for the clear proof and explanation sir.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Please help me finish this inductive proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 11th 2009, 07:06 PM
  2. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 10:33 PM
  3. Replies: 3
    Last Post: January 6th 2009, 09:34 PM
  4. how to finish this proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 2nd 2008, 12:40 PM
  5. Finish analysis Proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 14th 2007, 09:51 AM

Search Tags


/mathhelpforum @mathhelpforum