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Math Help - Geometric series

  1. #1
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    Geometric series

    Express the decimal as an rational number through geometric series

    1.213213213....

    \frac{ A} {1-r} \

    I understand how to do say .2222222222

    as \frac {2} {10}\ ( 1 + \frac { 1} {10}\ + \frac { 1} {100}\ ...)

    With my A = 2/10

    and my r = 1/10

    Any help would be appreciated! Im sure its me just brain farting, I cant seem to put it together.
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  2. #2
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    Here is a hint to get you started.
    0.213213\overline {213}  = \sum\limits_{n = 1}^\infty  {\frac{{213}}{{10^{3n} }}}
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  3. #3
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    Quote Originally Posted by Plato View Post
    Here is a hint to get you started.
    0.213213\overline {213}  = \sum\limits_{n = 1}^\infty  {\frac{{213}}{{10^{3n} }}}

    Hm still lost but i did manage to fudge an answer Algerbraically though not entirely correct

    by using An = 1.213213213

    then multiplying both sides by 100

    100An = 121.3213213

    then subtracting them and eventually getting


    \frac{ 120.108108} {99}\ = 1.213213213
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  4. #4
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    1.\overline {213}  = 1 + \sum\limits_{n = 1}^\infty  {\frac{{213}}{{10^{3n} }}}  = 1 + \frac{{\frac{{213}}{{1000}}}}{{1 - \frac{1}{{1000}}}} = 1 + \frac{{213}}{{999}}
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  5. #5
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    Quote Originally Posted by Plato View Post
    1.\overline {213}  = 1 + \sum\limits_{n = 1}^\infty  {\frac{{213}}{{10^{3n} }}}  = 1 + \frac{{\frac{{213}}{{1000}}}}{{1 - \frac{1}{{1000}}}} = 1 + \frac{{213}}{{999}}
    Thanks! I was not paying attention to the n=1 on the summation where on geometric series it doesn't effect the convergence/divergence just the value. Failing to realize that n=0 was my downfall. Case in point many thanks.
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