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Thread: fraction problems

  1. #1
    Newbie
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    Wink fraction problems

    Please help to solve these problems :

    1. (1+2/1) x (1+2/2) x (1+2/3) x .... x (1+2/26) x (1+2/27) =

    2. (1+1/4) x (1+1/9) x (1+1/16) x ... x (1+1/4028049) =

    Thank you
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  2. #2
    Senior Member Peritus's Avatar
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    (1+2/1) x (1+2/2) x (1+2/3) x .... x (1+2/26) x (1+2/27) = $\displaystyle \prod\limits_{k = 1}^{27} {\left( {1 + \frac{2}
    {k}} \right) = } \prod\limits_{k = 1}^{27} {\left( {\frac{{k + 2}}
    {k}} \right) = } \frac{{\frac{{29!}}
    {{2!}}}}
    {{27!}} = \frac{{28*29}}
    {2} = 14*29$

    2. (1+1/4) x (1+1/9) x (1+1/16) x ... x (1+1/4028049) = $\displaystyle \prod\limits_{k = 2}^{2007} {\left( {1 + \frac{1}
    {{k^2 }}} \right) = } \prod\limits_{k = 2}^{2007} {\left( {\frac{{1 + k^2 }}
    {{k^2 }}} \right) = } \frac{{\prod\limits_{k = 2}^{2007} {\left( {1 + k^2 } \right)} }}
    {{\left( {2007!} \right)^2 }}$
    let's denote the following product by:

    $\displaystyle
    P_m = \prod\limits_{k = 2}^m {\left( {1 + k^2 } \right)}
    $

    thus we can come up with the following recurrence relation for the product:

    $\displaystyle
    \left\{ \begin{gathered}
    P_{m + 1} = P_m \left[ {1 + \left( {m + 1} \right)^2 } \right] \hfill \\
    P_2 = 5 \hfill \\
    \end{gathered} \right.
    $

    so we can find $\displaystyle P_{2007} $ iteratively.

    $\displaystyle
    \text{(1 + 1/4) x (1 + 1/9) x (1 + 1/16) x }...\text{ x (1 + 1/4028049) = }\frac{{P_{2007} }}
    {{\left( {2007!} \right)^2 }}
    $
    Last edited by Peritus; Apr 16th 2008 at 06:04 AM.
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, agus hendro!

    Here's the first one . . .


    $\displaystyle 1)\;\; \;=\;\left(1+\frac{2}{1}\right)\cdot\left(1+\frac{ 2}{2}\right)\cdot\left(1+\frac{2}{3}\right) \hdots \left(1+\frac{2}{26}\right)\cdot\left(1+\frac{2}{2 7}\right)\; = $

    We have .$\displaystyle S \;=\;\frac{3}{1}\cdot\frac{4}{2}\cdot\frac{5}{3}\c dot\frac{6}{4}\cdot\frac{7}{5}\cdots\frac{28}{26}\ cdot\frac{29}{27}$ .$\displaystyle = \;\frac{\frac{29!}{2}}{27!}\;=\;\frac{29\cdot28}{2 } \;=\;\boxed{406}$

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  4. #4
    Newbie
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    Wink fraction problem correction

    The last two problem i've posted has been solved the first one. Thanks to Peritus and Soroban. But I should correct the secondsecond like this :
    2. (1 - 1/4) (1 - 1/9) (1 - 1/16) .... (1 - 1/2007^2).
    I think there is a simple solution like the first problem. Please help. Thank you
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  5. #5
    Senior Member Peritus's Avatar
    Joined
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    (1 - 1/4) (1 - 1/9) (1 - 1/16) .... (1 - 1/2007^2) = $\displaystyle \prod\limits_{k = 2}^{2007} {\left( {1 - \frac{1}
    {{k^2 }}} \right) = } \prod\limits_{k = 2}^{2007} {\left( {\frac{{k^2 - 1}}
    {{k^2 }}} \right) = } \prod\limits_{k = 2}^{2007} {\frac{{\left( {k - 1} \right)\left( {k + 1} \right)}}
    {{k^2 }} = } \frac{{2006!\frac{{2008!}}
    {2}}}
    {{\left( {2007!} \right)^2 }} = \frac{{2008}}
    {{2*2007}}$
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