Results 1 to 5 of 5

Math Help - Proof by mathematical induction.

  1. #1
    Member
    Joined
    Sep 2010
    Posts
    100

    Proof by mathematical induction.

    Show that the statement holds for all positive integers n:
    1^{2} + 2^{2}+...(n-1)^{3} < \frac{1}{4}n^{4} < 1^{3} +2^{3}+...n^{3}

    Firstly, I'll start with n = 1
    (1-1)^{3} < \frac{1}{4}(1)^{4} < (1)^{3}
    0 < \frac{1}{4} < 1
    Therefore, 1 ϵ S.
    Assuming k ϵ S:
    1^{2} + 2^{2}+...(k-1)^{3} < \frac{1}{4}k^{4} < 1^{3} +2^{3}+...k^{3}
    But, k ϵ S implies that k + 1 ϵ S, therefore:
    1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3}
    Simplify:
    2k^{3} - 3k^{2} + 3k - 1 < \frac{1}{4}k^{4} + k^{3} + \frac{3}{2}k^{2} + k + 1 < 2k^{3} + 3k^{2} + 3k + 1

    Therefore, k + 1 ϵ S and the statement holds for all positive integers n and all positive integers are in S.

    Have I proven this correctly by using mathematical induction? I haven't dealt with an equality yet in all the problem sets that I've done before. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2009
    Posts
    146

    Re: Proof by mathematical induction.

    1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3} is what you are trying to prove by induction, NOT something you already know to be true! You want to prove 1^{2} + 2^{2}+...(k-1)^{3} < \frac{1}{4}k^{4} < 1^{3} +2^{3}+...k^{3} \Longrightarrow 1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3}. How does it go from there?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    100

    Re: Proof by mathematical induction.

    Quote Originally Posted by godelproof View Post
    1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3} is what you are trying to prove by induction, NOT something you already know to be true! You want to prove 1^{2} + 2^{2}+...(k-1)^{3} < \frac{1}{4}k^{4} < 1^{3} +2^{3}+...k^{3} \Longrightarrow 1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3}. How does it go from there?
    That is where I seem to mess up. I haven't dealt with inequalities in my problem sets.

    So, I'll assume that:
    1^{2} + 2^{2}+...(k-1)^{3} = \frac{1}{4}(k+1)^{4} and 1^{3} +2^{3}+...k^{3} = (k-1)^{3}. Showing that is not true would prove validity of the original statement, correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member BAdhi's Avatar
    Joined
    Oct 2010
    From
    Gampaha, Sri Lanka
    Posts
    252
    Thanks
    6

    Re: Proof by mathematical induction.

    What I don't understand his how you explain the statement 1^2+2^2+...(n-1)^3. I mean from where does the power change from 2 to 3. Also I think that the statement should be written as 1^2+2^2+...+(n-1)^3 not 1^2+2^2+...(n-1)^3.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2010
    Posts
    100

    Re: Proof by mathematical induction.

    Quote Originally Posted by BAdhi View Post
    What I don't understand his how you explain the statement 1^2+2^2+...(n-1)^3. I mean from where does the power change from 2 to 3.
    That is how it asks in my book.

    Quote Originally Posted by BAdhi View Post
    Also I think that the statement should be written as 1^2+2^2+...+(n-1)^3 not 1^2+2^2+...(n-1)^3.
    Yes, you're correct with the notation. I made a typo.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mathematical Induction Proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 5th 2010, 11:24 AM
  2. Proof by Mathematical Induction
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: February 28th 2010, 08:07 AM
  3. Proof By Mathematical Induction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 6th 2008, 05:32 AM
  4. Replies: 5
    Last Post: October 19th 2007, 01:44 AM
  5. Proof Of Mathematical Induction
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: March 19th 2007, 08:24 PM

Search Tags


/mathhelpforum @mathhelpforum