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Math Help - A hard fraction problem

  1. #1
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    A hard fraction problem

    I have a problem with a fraction.

    (1+1/1)*(1+1/2)*(1+1/3)(1+1/99)

    How should i solve this?

    Thx in advance
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  2. #2
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    Re: A hard fraction problem

    Hint: express each factor as an improper fraction.
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  3. #3
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    Re: A hard fraction problem

    Can you give an example cuz I'm not good at these terms in english.
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    Re: A hard fraction problem

    Quote Originally Posted by Notatroll View Post
    Can you give an example cuz I'm not good at these terms in english.
    \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2} \right)\left(1+\frac{1}{3}\right)\cdots\left(1+ \frac{1}{99}\right)=\left(\frac{2}{1}\right)\left(  \frac{3}{2}\right)\left(\frac{4}{3}\right)\cdots \left(\frac{100}{99}\right)
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  5. #5
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    Re: A hard fraction problem

    Yeah I hae done that but how am I supposed to solve it? It's like 100000000 digits long :S
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  6. #6
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    Re: A hard fraction problem

    Oh thx. I just got it, isn't it f(x)=x+2? Every time you multiplice it it grows with 1, 98 times. Thx again to you Plato and emakarov!
    Last edited by Notatroll; August 19th 2012 at 06:27 AM. Reason: Forgot to thank the other person
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  7. #7
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    Re: A hard fraction problem

    Quote Originally Posted by Notatroll View Post
    Can you give an example cuz I'm not good at these terms in english.
    I also had to look it up in Wikipedia before I replied. Is it too difficult to Google it?

    Quote Originally Posted by Notatroll View Post
    Oh thx. I just got it, isn't it f(x)=x+2?
    I don't know what you mean by f(x).

    Quote Originally Posted by Notatroll View Post
    Every time you multiplice it it grows with 1, 98 times.
    This is also not clear. When you multiply, say (1 + 1/1)(1 + 1/2) by (1 + 1/3), it does not grow by 1.
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  8. #8
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    Re: A hard fraction problem

    Quote Originally Posted by Notatroll View Post
    Yeah I hae done that but how am I supposed to solve it? It's like 100000000 digits long :S
    \left( {\frac{{\rlap{/} 2}}{1}} \right)\left( {\frac{{\rlap{-/} 3}}{{\rlap{/} 2}}} \right)\left( {\frac{{\rlap{/} 4}}{{\rlap{/} 3}}} \right) \cdots \left( {\frac{{\rlap{/} 9\rlap{/} 9}}{{\rlap{/} 9\rlap{/} 8}}} \right)\left( {\frac{{100}}{{\rlap{/} 9\rlap{/} 9}}} \right) = 100
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  9. #9
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    Re: A hard fraction problem

    Rewrite 1 + \frac{1}{1} as \frac{2}{1}, 1 + \frac{1}{2} as \frac{3}{2} and so on. Pretty much everything cancels, as Plato showed. It essentially "telescopes."
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  10. #10
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    Re: A hard fraction problem

    Ehh... (2/1)*(3/2)*(4/3) is actually 4, so it does grow with 1 all the time.. So i suppose i was correct.
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