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Math Help - fraction problems

  1. #1
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    Smile fraction problems

    Please help answer these problems

    1.find the value of 1/(4x9) + 1/(9x14) + 1/(14x19) + ........ + 1/(1999x2004)

    2.find the value of 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + ....... + 1/300

    I'm waiting..... Thank you
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  2. #2
    Junior Member teuthid's Avatar
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    what level of math did this problem come from? Would it make sense to you if I told you that the first sum could be expressed as \displaystyle\sum_{n=1}^{400}\frac{1}{(5n-1)(5n+4)}?
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  3. #3
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    Unhappy

    This problem was one of the problems in Math Olympiad for the elementary school. The time allocated for the problems is about 5 minutes. So there must be simple but logical answer. Please help...
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  4. #4
    Junior Member teuthid's Avatar
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    Sorry, I can't think of anything that could be done by an elementary student in 5 minutes.
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  5. #5
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    Hello, cdesiani!

    Elementary school?
    Perhaps very bright elementary students who have mastered Algebra
    . . and know some clever (sneaky) tricks . . .


    1) Find the value of: . \frac{1}{4\!\cdot\!9} + \frac{1}{9\!\cdot\!14} + \frac{1}{14\!\cdot\!19} + \cdots + \frac{1}{1999\!\cdot\!2004}

    We note that: . \begin{Bmatrix}\dfrac{1}{4 \cdot 9} &=&\dfrac{1}{5}\left(\dfrac{1}{4} - \dfrac{1}{9}\right) \\ \\[-2mm]<br />
\dfrac{1}{9\cdot14} &=&\dfrac{1}{5}\left(\dfrac{1}{9} - \dfrac{1}{14}\right) \\  \\[-2mm]<br />
\dfrac{1}{14\cdot9} &=&\dfrac{1}{5}\left(\dfrac{1}{14} - \dfrac{1}{19}\right) \\ \vdots & & \vdots \end{Bmatrix}


    We have: . \frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}\right) + \frac{1}{5}\left(\frac{1}{9}-\frac{1}{14}\right) + \frac{1}{5}\left(\frac{1}{14} - \frac{1}{19}\right) + \hdots + \frac{1}{5}\left(\frac{1}{1999} - \frac{1}{2004}\right)

    . . . . . . = \;\frac{1}{5}\left(\frac{1}{4} - \frac{1}{9} + \frac{1}{9} - \frac{1}{14} + \frac{1}{14} - \frac{1}{19} + \hdots + \frac{1}{1999} - \frac{1}{2004}\right)

    . . . . . . = \;\frac{1}{5}\left(\frac{1}{4} - \frac{1}{2004}\right) \;=\;\frac{1}{5}\left(\frac{500}{2004}\right) \;=\;\boxed{\frac{25}{501}}



    2) Find the value of: . \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \cdots + \frac{1}{300}

    We note that: . \begin{Bmatrix}3 &=&\dfrac{2\cdot3}{2} \\ \\[-2mm]<br />
6 &=&\dfrac{3\cdot4}{2} \\ \\[-2mm]<br />
10 &=&\dfrac{4\cdot5}{2} \\ \\[-2mm]<br />
15 &=&\dfrac{5\cdot6}{2} \\ & \vdots \end{Bmatrix}


    We have: . \frac{1}{\frac{2\cdot3}{2}} + \frac{1}{\frac{3\cdot4}{2}} + \frac{1}{\frac{4\cdot5}{2}} + \frac{1}{\frac{5\cdot6}{2}} + \hdots + \frac{1}{\frac{24\cdot25}{2}} . . . . . = \;\frac{2}{2\cdot3} + \frac{2}{3\cdot4} + \frac{2}{4\cdot5} + \frac{2}{5\cdot6} + \hdots + \frac{2}{24\cdot25}

    . . . . . = \;2\left[\frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \frac{1}{4\cdot5} + \frac{1}{5\cdot6} + \hdots + \frac{1}{24\cdot25}\right]

    . . . . . = \;2\left[\left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \hdots + \left(\frac{1}{24} - \frac{1}{25}\right)\right]

    . . . . . = \;2\left(\frac{1}{2}-\frac{1}{25}\right) \;=\;2\left(\frac{23}{50}\right) \;=\;\boxed{\frac{23}{25}}



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  6. #6
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    Talking Thank you, it's solved!!

    Wow, what a beautiful way to solve such a difficult problem!

    Thank you so much!
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  7. #7
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    WELL........
    I CAN NOT HELP U I DO NOT NO HOW 2 DO IT ETHER !!!!

    *****SORRY!!!!!!!!!!*****
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  8. #8
    Moo
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    Hi,

    Please be nice with my eyes
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