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Math Help - Binomial Theorem

  1. #1
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    Binomial Theorem

    Hi again guys,

    Sorry to bother you again, i have another question which i am totally unfamiliar with. If anybody could guide me through the solution it would be great.

    Ok here goes...

    Questions:

    A) Using the technique of the Sum to infinity express as a fraction in its lowest terms the following recurring decimal number

    0.05 (recurring)




    B) Using the Binomial Theorem find the first 5 terms of the Binomial expression

    (1 + 0.2)^12



    ??? GULP ???
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  2. #2
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    The binomial thereom:

    (1+x)^{m}=1+mx+\frac{m(m-1)}{2!}x^{2}+ ......+\frac{m(m-1)(m-2).......(m-k+1)}{k!}x^{k}

    For yours:

     1+12x+\frac{12(11)}{2!}x^{2}+\frac{12(11)(10)}{3!}  x^{3}+....etc

    Then use x=.02
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  3. #3
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    Ok part B i got 8.84544 after i used the theorem.

    Can anybody help with part A?
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  4. #4
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    Quote Originally Posted by c00ky

    A) Using the technique of the Sum to infinity express as a fraction in its lowest terms the following recurring decimal number

    0.05 (recurring)

    I presume you mean,
    .0505050505050....
    Then, by definition of decimals,
    \frac{5}{10^2}+\frac{5}{10^4}+\frac{5}{10^6}+...
    Factor, (you can manipulate this series anyway you like since it is absolutely convergent)
    \frac{5}{10^2}\left( 1+\frac{1}{10^2}+\frac{1}{10^4}+... \right)
    Express as sum,
    \frac{5}{100}\sum_{k=0}^{\infty} (100)^{-k}
    Use geometry sum,
    \frac{5}{100}\cdot \frac{1}{1-\frac{1}{100}}
    Simplify,
    \frac{5}{100}\cdot \frac{100}{100-1}=\frac{5}{99}
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  5. #5
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    Hello, c00ky!

    Here's part (A) . . .


    A) Using the technique of the Sum to infinity, express as a fraction in its lowest terms
    the following recurring decimal number: .0.05 (recurring)

    It's not clear what part is recurring.
    . . I'll assume that the number is: . N \:= \:0.0555...

    We have: . N\;=\;\frac{5}{100} + \frac{5}{1000} + \frac{5}{10,000} + \frac{5}{100,000} + \hdots

    Factor: . . N \;= \;\frac{5}{100}\underbrace{\left(1 + \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \hdots\right)}

    We have a geometric series with a = 1,\;r = \frac{1}{10}
    . . Its sum is: . S\:=\:\frac{1}{1 - \frac{1}{10}} \:=\:\frac{10}{9}


    Therefore: . N\;=\;\left(\frac{5}{100}\right)\left(\frac{10}{9}  \right)\;= \;\boxed{\frac{1}{18}}

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  6. #6
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    A BIG thanks to everybody. Soroban that's the part which is recurring, that's great thanks.

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