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Thread: summation and mathematical induction

  1. #1
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    summation and mathematical induction

    1.Prove by induction that $\displaystyle \sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)}) $$\displaystyle =$$\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}$.

    Show that $\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}$ is LESSER than $\displaystyle \frac{1}{4}$ for all positive integer values of n.

    Deduce from these results that $\displaystyle \sum_{r=1}^{n}(\frac{1}{(r+1)^3})$ is LESSER than $\displaystyle \frac{1}{4}$.


    i can do the first part of the question. i need help for the second and third parts in getting started.
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  2. #2
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    Quote Originally Posted by wintersoltice View Post
    1.Prove by induction that $\displaystyle \sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)}) $$\displaystyle =$$\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}$.

    Show that $\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}$ is LESSER than $\displaystyle \frac{1}{4}$ for all positive integer values of n.

    Deduce from these results that $\displaystyle \sum_{r=1}^{n}(\frac{1}{(r+1)^3})$ is LESSER than $\displaystyle \frac{1}{4}$.


    i can do the first part of the question. i need help for the second and third parts in getting started.
    For the second part

    $\displaystyle \frac{1}{4}\left(\frac{n(n+3)}{(n+1)(n+2)}\right)< \frac{1}{4}$ if n(n+3)<(n+1)(n+2)

    $\displaystyle n^2+3n<n^2+3n+2\ ?$ yes, it is
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  3. #3
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    Quote Originally Posted by wintersoltice View Post
    1.Prove by induction that $\displaystyle \sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)}) $$\displaystyle =$$\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}$.

    Show that $\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}$ is LESSER than $\displaystyle \frac{1}{4}$ for all positive integer values of n.

    Deduce from these results that $\displaystyle \sum_{r=1}^{n}(\frac{1}{(r+1)^3})$ is LESSER than $\displaystyle \frac{1}{4}$.


    i can do the first part of the question. i need help for the second and third parts in getting started.
    The third part involves showing that

    $\displaystyle \frac{1}{(r+1)^3}<\frac{1}{r(r+1)(r+2)}$

    $\displaystyle \frac{1}{(r+1)^2}<\frac{1}{r(r+2)}\ ?$

    $\displaystyle \frac{1}{r^2+2r+1}<\frac{1}{r^2+2r}\ ?$

    $\displaystyle r^2+2r+1>r^2+2r\ ?$


    The induction proof appears more challenging than parts 2 and 3.
    Here's how I would give the Induction Proof....

    P(k)

    $\displaystyle \sum_{r=1}^{k}\frac{1}{r(r+1)(r+2)}=\frac{k(k+3)}{ 4(k+1)(k+2)}\ ?$

    P(k+1)

    $\displaystyle \sum_{r=1}^{k+1}\frac{1}{r(r+1)(r+2)}=\frac{(k+1)( k+4)}{4(k+2)(k+3)}\ ?$

    Try to prove that P(k) being true causes P(k+1) to also be true.

    Proof

    $\displaystyle \sum_{r=1}^{k+1}\frac{1}{r(r+1)(r+2)}=\sum_{r=1}^{ k}\frac{1}{r(r+1)(r+2)}+\frac{1}{(k+1)(k+2)(k+3)}$

    $\displaystyle =\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+ 3)}$

    if P(k) is true.

    This gives

    $\displaystyle \frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}+\frac{4}{k+1}\le ft(\frac{1}{4(k+2)(k+3)}\right)$

    $\displaystyle \frac{1}{4(k+2)(k+3)}\left(\frac{k(k+3)^2}{k+1}+\f rac{4}{k+1}\right)$

    $\displaystyle =\frac{1}{4(k+2)(k+3)}\left(\frac{k\left(k^2+6k+9\ right)+4}{k+1}\right)$

    $\displaystyle =\frac{1}{4(k+2)(k+3)}\left(\frac{k^3+6k^2+9k+4}{k +1}\right)$

    $\displaystyle =\frac{1}{4(k+2)(k+3)}\frac{(k+1)\left(k^2+5k+4\ri ght)}{k+1}$

    $\displaystyle =\frac{(k+4)(k+1)}{4(k+2)(k+3)}$

    Then test your initial value to complete the proof.
    Last edited by Archie Meade; Jul 15th 2010 at 08:58 AM.
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