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Thread: equality problem with fractions

  1. #1
    Newbie
    Joined
    Apr 2010
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    andalusia
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    equality problem with fractions

    hi. nice to be here.
    I have to show, that

    $\displaystyle
    \frac{n}{2n+1}
    +
    \frac{1}{(2n+1)*(2n+3)}
    =
    \frac{n+1}{2n+3}
    $

    basicly i must confess, that i don't remember quiet a lot from school and that seems to be my problem with this. I was looking for a way to simplyfy this and my only idea was to add the two fractions. my result was:

    $\displaystyle
    \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)}
    =
    $
    $\displaystyle \frac{n+1}{2n+3}
    $

    which -i believe- is the same as:

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{2*(2n+1)*(2n+3)}
    $

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{(4n+2)*(2n+3)}
    $

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{8n^2+12n+4n+6}
    $

    $\displaystyle
    =\frac{8n^3+12n^2+2n^2+3n+2n+1}{8n^2+16n+6}
    $

    $\displaystyle
    =\frac{8n^3+14n^2+5n+1}{8n^2+16n+6}
    $


    ...and basicly this is status quo, because this is, with what i have spent the last hours. i feel pleased for any answers or recommentations: this might be a trivial problem for some of you, but not for me! It's a shame, that i probably have problems with some very basic mathematical operations. please help me!
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  2. #2
    MHF Contributor undefined's Avatar
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    Chicago
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    Quote Originally Posted by Maya View Post
    hi. nice to be here.
    I have to show, that

    $\displaystyle
    \frac{n}{2n+1}
    +
    \frac{1}{(2n+1)*(2n+3)}
    =
    \frac{n+1}{2n+3}
    $

    basicly i must confess, that i don't remember quiet a lot from school and that seems to be my problem with this. I was looking for a way to simplyfy this and my only idea was to add the two fractions. my result was:

    $\displaystyle
    \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)}
    =
    $
    $\displaystyle \frac{n+1}{2n+3}
    $

    which -i believe- is the same as:

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{2*(2n+1)*(2n+3)}
    $

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{(4n+2)*(2n+3)}
    $

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{8n^2+12n+4n+6}
    $

    $\displaystyle
    =\frac{8n^3+12n^2+2n^2+3n+2n+1}{8n^2+16n+6}
    $

    $\displaystyle
    =\frac{8n^3+14n^2+5n+1}{8n^2+16n+6}
    $


    ...and basicly this is status quo, because this is, with what i have spent the last hours. i feel pleased for any answers or recommentations: this might be a trivial problem for some of you, but not for me! It's a shame, that i probably have problems with some very basic mathematical operations. please help me!
    Without looking too closely at your work, the first step in your addition of fractions could be made a lot simpler by recognizing that the LCM of (2n+1)*(2n+3) and 2n+1 is (2n+1)*(2n+3), not (2n+1)*(2n+1)*(2n+3).

    Hope that helps.
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  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
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    Germany
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    Quote Originally Posted by Maya View Post
    hi. nice to be here.
    I have to show, that

    $\displaystyle
    \frac{n}{2n+1}
    +
    \frac{1}{(2n+1)*(2n+3)}
    =
    \frac{n+1}{2n+3}
    $

    basicly i must confess, that i don't remember quiet a lot from school and that seems to be my problem with this. I was looking for a way to simplyfy this and my only idea was to add the two fractions. my result was:

    $\displaystyle
    \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)}
    =
    $
    $\displaystyle \frac{n+1}{2n+3}
    $

    which -i believe- is the same as:

    $\displaystyle
    =\frac{(\bold{\color{red}2}n^2+1n)*(2n+3)+2n+1}{2* (2n+1)*(2n+3)}
    $ <<<<<<<< typo

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{(4n+2)*(2n+3)}
    $

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{8n^2+12n+4n+6}
    $

    $\displaystyle
    =\frac{8n^3+12n^2+2n^2+3n+2n+1}{8n^2+16n+6}
    $

    $\displaystyle
    =\frac{8n^3+14n^2+5n+1}{8n^2+16n+6}
    $


    ...and basicly this is status quo, because this is, with what i have spent the last hours. i feel pleased for any answers or recommentations: this might be a trivial problem for some of you, but not for me! It's a shame, that i probably have problems with some very basic mathematical operations. please help me!
    If I were you I would factor out $\displaystyle \frac1{2n+1}$ :

    $\displaystyle \frac{n}{2n+1} + \frac{1}{(2n+1)*(2n+3)} = \frac1{2n+1}\left(n+\frac1{2n+3}\right)$

    Now simplify the term in the bracket. Keep in mind that $\displaystyle 2n^2+3n+1=(n+1)(2n+1)$

    Cancel the factor $\displaystyle (2n+1)$ to get the given result.
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  4. #4
    Newbie
    Joined
    Apr 2010
    From
    andalusia
    Posts
    5
    I really tried hard, but this really starts to make me sad, because i am not able to understand your answers. my problem seems to be more fundamental than i ever thought. thanks for your advise!

    top 1

    As I understand the hint, that "the LCM of (2n+1)*(2n+3) and 2n+1 is (2n+1)*(2n+3), not (2n+1)*(2n+1)*(2n+3)":

    Let me assume, that

    $\displaystyle
    \frac{a}{b}+\frac{c}{d}
    =
    \frac{a*d+c*b}{\color{red}{b*d}}
    $

    so, that for

    $\displaystyle
    \frac{n}{2n+1}
    +
    \frac{1}{(2n+1)*(2n+3)}
    $

    i get

    $\displaystyle
    a := n
    $

    $\displaystyle
    b := (2n+1)
    $

    $\displaystyle
    c := 1
    $

    $\displaystyle
    d := (2n+1)*(2n+3)
    $

    which leads me to:

    $\displaystyle
    \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)}
    $

    and not

    $\displaystyle
    \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+3)}
    $

    what's wrong about?


    top 2
    i could not find a good source, which describes how to factor out something works yet. I believe this is, why i cannot see why

    $\displaystyle
    \frac{n+1}{2n+3}
    =
    \frac1{2n+1}\left(n+\frac1{2n+3}\right)
    $
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  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
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    1
    Quote Originally Posted by Maya View Post
    hi. nice to be here.
    I have to show, that

    $\displaystyle
    \frac{n}{2n+1}
    +
    \frac{1}{(2n+1)*(2n+3)}
    =
    \frac{n+1}{2n+3}
    $

    basicly i must confess, that i don't remember quiet a lot from school and that seems to be my problem with this. I was looking for a way to simplyfy this and my only idea was to add the two fractions. my result was:

    $\displaystyle
    \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)}
    =
    $
    $\displaystyle \frac{n+1}{2n+3}
    $

    which -i believe- is the same as:

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{2*(2n+1)*(2n+3)}
    $

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{(4n+2)*(2n+3)}
    $

    $\displaystyle
    =\frac{(4n^2+1n)*(2n+3)+2n+1}{8n^2+12n+4n+6}
    $

    $\displaystyle
    =\frac{8n^3+12n^2+2n^2+3n+2n+1}{8n^2+16n+6}
    $

    $\displaystyle
    =\frac{8n^3+14n^2+5n+1}{8n^2+16n+6}
    $


    ...and basicly this is status quo, because this is, with what i have spent the last hours. i feel pleased for any answers or recommentations: this might be a trivial problem for some of you, but not for me! It's a shame, that i probably have problems with some very basic mathematical operations. please help me!
    Multiply $\displaystyle \frac{n}{2n+1}$ by $\displaystyle \frac{2n+3}{2n+3}$.

    This gives

    $\displaystyle \frac{n(2n+3)}{(2n+1)(2n+3)} + \frac{1}{(2n+1)(2n+3)}$

    These have the same denominator so you can add them

    See the spoiler for why you need only multiply by $\displaystyle \frac{2n+3}{2n+3}$
    Spoiler:

    Consider $\displaystyle \frac{a}{b} + \frac{c}{D}$

    For this special case $\displaystyle D = bd$

    Hence the sum is \frac{a}{b}+\frac{c}{bd}

    If we cross multipy: $\displaystyle \frac{abd+bc}{bd} = \frac{b(ad+bc)}{bd}$. Note that b will cancel and it was as though we just multiplied $\displaystyle \frac{a}{b} \cdot \frac{d}{d}$

    (this is what undefined was pointing to with LCD)
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