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Math Help - Induction Problem

  1. #1
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    Induction Problem

    Hi I am having alot of trouble with this induction question. Any help would be much appreciated!

    Find a formula for 1/1*2 + 1/2*3 + 1/n*(n+1) and prove by induction.
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  2. #2
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    Hi

    To find the formula you can compute the first terms of the series.
    Another way is to use
    \frac{1}{n(n+1)}=\frac{1}{n} - \frac{1}{n+1}
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  3. #3
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    So could the formula be
    \frac{1}{n} - \frac{1}{n+1} or is this just another way used to solve ?
    I can calculate the first few terms but I am not able to find a formula.
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  4. #4
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    Quote Originally Posted by vexiked View Post
    So could the formula be
    \frac{1}{n} - \frac{1}{n+1} or is this just another way used to solve ?
    This is just a way to solve

    Quote Originally Posted by vexiked View Post
    I can calculate the first few terms but I am not able to find a formula.
    You should : what are your results ?
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  5. #5
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    Quote Originally Posted by vexiked View Post
    Hi I am having alot of trouble with this induction question. Any help would be much appreciated!

    Find a formula for 1/1*2 + 1/2*3 + 1/n*(n+1) and prove by induction.
    Here's some details. You'll notice that each term can be split up

    \frac{1}{1 \cdot 2} = \frac{1}{1} - \frac{1}{2},\;\;\;\;\;\frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3},\;\;\;\;\;\frac{1}{3 \cdot 4} = \frac{1}{3} - \frac{1}{4}
    \frac{1}{n \cdot (n+1)} = \frac{1}{n} - \frac{1}{n+1}
    Then when you add them up all the terms cancel except the first and last, so

    S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}

    For the induction part, it true for the first few, assume true for n = k

    so

    S_k = \frac{k}{k+1}

    Prove true for n = k+1, i.e. S_{k+1} = \frac{k+1}{k+2}
    so


    S_{k+1} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)}

    S_{k+1} = \frac{k+1}{k+2}
    so it must be true for all n.
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  6. #6
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    Could you verify that I am on the right track for another?

    Prove that 3^n < n! if n is an integer greater than 6..

    - Base case 3^7 = 2187 and 7! = 5040
    - Inductive
    - Assume for p(k) show for p(k+1)
    - 3^K+1 < K+1!
    - 3^k + 3^k < K + 1!

    I am stuck on this part now.
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  7. #7
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    You should have an initial condition that n>2

     <br />
3^{n} < n!<br />
    Let the following is true
    P(k) = 3^{k}<k!
    To prove
     <br />
P(k+1) = 3^{k+1}<(k+1)!<br />
    Therefore

     <br />
3^{k}\times  3 < k!(k+1)<br />
    Putting k! instead of 3^k we get something greater hence if we can prove that this greater thing less than our RHS
    then we have done it
     <br />
3*k! < k! (k+1) <br /> <br />
    For  k > 2

    Thus 3<(k+1)!
    and  <br />
3k!<(k+1)!<br /> <br />
    Last edited by ADARSH; February 8th 2009 at 08:28 AM. Reason: added
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  8. #8
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    Quote Originally Posted by running-gag View Post
    Hi

    To find the formula you can compute the first terms of the series.
    Another way is to use
    \frac{1}{n(n+1)}=\frac{1}{n} - \frac{1}{n+1}

    Could someone show me how to compute the first terms of the series to get the formula? Im in dire need, have an exam tomorrow and I have had no luck figuring out the steps.
    Last edited by vexiked; April 19th 2009 at 08:39 PM.
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  9. #9
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    danny arrigo already answered to you
    What don't you understand ?
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