# Series

• Apr 12th 2014, 07:57 AM
srirahulan
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)
• Apr 12th 2014, 08:47 AM
Plato
Re: Series
Quote:

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

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

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?
• Apr 12th 2014, 11:19 AM
Soroban
Re: Series
Hello, srirahulan!

Quote:

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

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

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

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

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

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

. . . . . . . . .$\displaystyle S \;=\;4 - T_n$
• Apr 12th 2014, 07:22 PM
srirahulan
Re: Series
Thank you so much>....>>