Results 1 to 4 of 4

Thread: Series

  1. #1
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    180

    Lightbulb Series

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

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

    Show that
    $\displaystyle S_{n}=4-T_{n}$
    In this case how can i do this ? can i find the $\displaystyle 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; Apr 12th 2014 at 08:03 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,734
    Thanks
    2810
    Awards
    1

    Re: Series

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

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

    Show that
    $\displaystyle S_{n}=4-T_{n}$
    In this case how can i do this ? can i find the $\displaystyle 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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848

    Re: Series

    Hello, srirahulan!

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

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

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

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

    $\displaystyle \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 $\displaystyle a = 1$, common ratio $\displaystyle r = \tfrac{1}{2}$, and $\displaystyle n$ terms.
    . . Its sum is: .$\displaystyle \frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}} \;=\;2-2(\tfrac{1}{2})^n $

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

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

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

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

    . . . . . . . . .$\displaystyle S \;=\;4 - T_n$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    180

    Re: Series

    Thank you so much>....>>
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Dec 9th 2012, 12:50 PM
  2. Replies: 2
    Last Post: May 22nd 2012, 05:57 AM
  3. Replies: 3
    Last Post: Sep 29th 2010, 06:11 AM
  4. Replies: 0
    Last Post: Jan 26th 2010, 08:06 AM
  5. Replies: 1
    Last Post: May 5th 2008, 09:44 PM

Search Tags


/mathhelpforum @mathhelpforum