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Math Help - Series

  1. #1
    Member srirahulan's Avatar
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    Lightbulb Series

    S_{n}=1+2\frac{1}{2}+3\frac{1}{2}^{2}+4\frac{1}{2}  ^{3}+.............+n\frac{1}{2}^{n-1}

    T_{n}=(n+2)\frac{1}{2}^{n-1} n-Z^{+}

    Show that
    S_{n}=4-T_{n}
    In this case how can i do this ? can i find the U_{r} of this series and then continue???? plz help>>>

    note--i cannot apply the hole bracket for that (latex problem)
    Last edited by srirahulan; April 12th 2014 at 08:03 AM.
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  2. #2
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    Re: Series

    Quote Originally Posted by srirahulan View Post
    S_{n}=1+2\frac{1}{2}+3\frac{1}{2}^{2}+4\frac{1}{2}  ^{3}+.............+n\frac{1}{2}^{n-1}

    T_{n}=(n+2)\frac{1}{2}^{n-1} n-Z^{+}

    Show that
    S_{n}=4-T_{n}
    In this case how can i do this ? can i find the U_{r} of this series and then continue???? plz help>>>

    note--i cannot apply the hole bracket for that (latex problem)
    I have no idea of how that series is defined. Is it ${S_n} = \sum\limits_{k = 1}^n {\dfrac{k}{{{2^{k - 1}}}}}~? $

    The laTex is \$\dfrac{k}{2^{-k}}\$ so what does "note--i cannot apply the hole bracket for that" mean?
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  3. #3
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    Re: Series

    Hello, srirahulan!

    S_n \;=\;1+2(\tfrac{1}{2})+3(\tfrac{1}{2})^2+4(\tfrac{  1}{2})^3+ \cdots + n(\tfrac{1}{2})^{n-1}

    T_n\;=\;(n+2)(\tfrac{1}{2})^{n-1}\;\;\; n \in Z^+

    Show that: . S_n \;=\;4-T_n

    \begin{array}{ccccc}\text{We have:} & S &=& 1+2(\tfrac{1}{2})+3(\tfrac{1}{2})^2+4(\tfrac{1}{2}  )^3+ \cdots + n(\tfrac{1}{2})^{n-1} \qquad\qquad\qquad \\ \text{Multiply by }\frac{1}{2}: & \frac{1}{2}S &=& \qquad\;\;\; \frac{1}{2}+2(\tfrac{1}{2})^2+3(\tfrac{1}{2})^3+  \cdots + (n-1)(\tfrac{1}{2})^{n-1} + n(\frac{1}{2})n  \\ \text{Subtract:} & \tfrac{1}{2}S &=& 1 + \tfrac{1}{2} + (\tfrac{1}{2})^2 + (\tfrac{1}{2})^3 + \cdots + (\tfrac{1}{2})^{n-1} - n(\tfrac{1}{2})^n  \end{array}

    \text{We have: }\;\tfrac{1}{2}S \;=\; \underbrace{1 + \tfrac{1}{2} + (\tfrac{1}{2})^2 + (\tfrac{1}{2})^3 + \cdots + (\tfrac{1}{2})^{n-1}}_{\text{geometric series}} - n(\tfrac{1}{2})^n

    The series has first term a = 1, common ratio r = \tfrac{1}{2}, and n terms.
    . . Its sum is: . \frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}} \;=\;2-2(\tfrac{1}{2})^n

    We have: . \tfrac{1}{2}S \;=\;2 - 2(\tfrac{1}{2})^n - n(\tfrac{1}{2})^n

    . . . . . . . . \tfrac{1}{2}S \;=\;2 - (n+2)(\tfrac{1}{2})^n

    . . . . . . . . . S \;=\;4 - 2(n+2)(\tfrac{1}{2})^n

    . . . . . . . . . S \;=\;4 - (n+2)(\tfrac{1}{2})^{n-1}

    . . . . . . . . . S \;=\;4 - T_n
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  4. #4
    Member srirahulan's Avatar
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    Re: Series

    Thank you so much>....>>
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