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Math Help - Need help with Mathematical induction question?

  1. #1
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    Need help with Mathematical induction question?

    I'm working on a problem and I'm not sure I have this correct on how to do it.
    here is the problem
    Prove the following statement by mathematical induction:
    1/2+1/6+...+1/n(n+1)=n/n+1, for all integers n> or = 1
    My first thought is to solve the n/n+1 side with the number 1
    So I get 1/2 on both sides by substituting. Then I start to become unsure on where to go from there.
    Any help would be awesome.
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  2. #2
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    Re: Need help with Mathematical induction question?

    First show the base case P_1 is true:

    \frac{1}{2}=\frac{1}{1(1+1)}=\frac{1}{2}

    True.

    State the induction hypothesis P_n:

    \sum_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}

    Add to this the equation:

    \frac{1}{(n+1)((n+1)+1)}=\frac{1}{(n+1)(n+2)}

    Simplify by incorporating the added term within the summation on the left and combining terms on the right, and you will find you have derived P_{n+1} from P_n, thereby completing the proof by induction.
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  3. #3
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    Re: Need help with Mathematical induction question?

    Induction is not necessary.

    \displaystyle \begin{align*} \sum_{k = 1}^n {\frac{1}{k(k + 1)}} &= \sum_{k = 1}^n{\left( \frac{1}{k} - \frac{1}{k + 1} \right)} \\ &= \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n} + \frac{1}{n + 1} \right) \\ &= 1 - \frac{1}{n + 1} \\ &= \frac{n}{n + 1} \end{align*}
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  4. #4
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    Re: Need help with Mathematical induction question?

    Mark, awesome answer, but a quick question I thought you had to go prove it in terms of K+1. Although I've been wrong before, thanks a bunch.
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Need help with Mathematical induction question?

    If you carry out the operations I suggested you will get:

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

    This is P_{n+1}, which was derived from P_n.
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  6. #6
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    Re: Need help with Mathematical induction question?

    Wait, so how did we get 1/(n+1)((n+1)+1) = n+1/n+2 sorry for drawing this question out just trying to understand the different parts.
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: Need help with Mathematical induction question?

    State the induction hypothesis P_n:

    \sum_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}

    Add to this the equation:

    \frac{1}{(n+1)((n+1)+1)}=\frac{1}{(n+1)(n+2)}

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

    On the left we may incorporate the added term within the summation:

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

    On the right combine terms:

    \sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{1}{n+1} \left(n+\frac{1}{n+2} \right)

    \sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{1}{n+1} \left(\frac{n(n+2)+1}{n+2} \right)

    \sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{1}{n+1} \left(\frac{n^2+2n+1}{n+2} \right)

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

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

    \sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{n+1}{(n+1)+  1}

    Now we have P_{n+1}, completing the proof by induction.
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