Results 1 to 5 of 5

Math Help - Induction Help

  1. #1
    Newbie
    Joined
    Dec 2012
    From
    USA
    Posts
    2

    Induction Help

    ∑ i=1 to n, √[1+(1/i^2)+(1/(1+i^2))] = n(n+2)/n+1


    First I did the base case of p(1) showing 3/2 on the LHS equals the 3/2 on the RHS.
    Then I assumed p(k) and wrote out the formula with k in it.
    Then prove p(k+1)= p(k)+ √[1+1/(k+1)^2+1/(k+2)^2]
    =k(k+2)/k+1 + √[1+1/(k+1)^2+1/(k+2)^2]
    Then I squared each to get rid of the square root.
    (k(k+2)/(k+1))^2+ (k+1)^2/(k+1)^2 + 1/(k+1)^2 + 1/(k+2)^2
    Now I'm stuck any Guidance would be great thanks!
    Last edited by andyk23; December 10th 2012 at 07:29 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1

    Re: Induction Help

    There are a couple of problems with your post. Firstly I don't believe that formula is correct. According to your formula the LHS when n=1 is \sqrt{1+1+\frac{1}{2}}=\frac{\sqrt{10}}{2} \neq \frac{3}{2}.
    I believe what you should be trying to prove is
    \sum_{i=1}^{n}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)  ^2}}=\frac{n(n+2)}{n+1}.

    Secondly your approach for taking the inductive step seems a little strange- in such proofs you prove the base case (i.e. n=1) and then assume the formula to be true for n=1,..,k. Then you consider the LHS when n=k+1, and use the fact that
    \sum_{i=1}^{k+1}f(i)=\sum_{i=1}^kf(i)+f(k+1)
    to prove the formula for all n.

    I hope this helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2012
    From
    USA
    Posts
    2

    Re: Induction Help

    I'm just having trouble showing that, \frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{k+1^2}+\frac{1}{(1+k+1)^2}}=\frac  {k+1(k+1+2)}{k+1+1}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1

    Re: Induction Help

    You've got to mess around a bit with the algebra:

    \sqrt{1+\frac{1}{i^2}+\frac{1}{(i+1)^2}}=\frac{1}{  i(i+1)}\sqrt{i^2(i+1)^2+(i+1)^2+i^2},

    Multiplying out the term in the square root and collecting like terms gives

    \frac{1}{i(i+1)}\sqrt{i^4+2i^3+3i^2+2i+1}=\frac{1}  {i(i+1)}\sqrt{(i^2+i+1)^2}=\frac{i^2+i+1}{i(i+1)}

    The right hand side can then be rewritten as

    g(i)=\frac{i^2(i+2)-(i-1)(i+1)^2}{i(i+1)}.

    Now using g(i) instead of \sqrt{1+\frac{1}{i^2}+\frac{1}{(i+1)^2}} for i=k+1 you should be able to complete the induction.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Induction Help

    You might want to take a look at this topic:

    Sum of a finite series.

    for a way to simplify your induction hypothesis.

    edit: If I had taken the time to actually read the post above this one before posting, I would have seen the same simplification has already been given.
    Last edited by MarkFL; December 12th 2012 at 03:02 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Strong induction vs. structural induction?
    Posted in the Discrete Math Forum
    Replies: 13
    Last Post: April 21st 2011, 12:36 AM
  2. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  3. need help with Induction
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 18th 2010, 11:39 AM
  4. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 09:33 PM
  5. Induction
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: October 5th 2007, 04:22 PM

Search Tags


/mathhelpforum @mathhelpforum