Results 1 to 9 of 9

Thread: Sum to n of series

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Sum to n of series

    If $\displaystyle S_n=\displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)$, $\displaystyle (|a|\neq 1)$ by considering $\displaystyle (1-a)S_n$, show that
    $\displaystyle S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
    State the set of values of a for which $\displaystyle S_n$approaches a limit as $\displaystyle n \rightarrow\infty$ and find the sum to infinity of the series for these values.

    Attempt:
    $\displaystyle \displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)$
    $\displaystyle =\displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^n_{r=0}a^r\right)$
    $\displaystyle =\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^r}{1-a}\right)$
    $\displaystyle =\displaystyle \sum^n_{r=0}\frac{a^r-a^{2r}}{1-a}$
    Am I correct so far? and how to continue because I don't see a way of summing that?
    Thanks!
    Last edited by arze; Aug 17th 2010 at 04:04 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    This is almost correct. Here is the correct version:

    $\displaystyle \displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^r_{k=0}a^k\right)$

    $\displaystyle =\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^{r+1}}{1-a}\right)$

    In the question you are told to consider $\displaystyle (1-a)S_n$. Why don't you try that!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    $\displaystyle (1-a)S_n=S_n-aS_n$
    $\displaystyle =\displaystyle \sum^n_{r=0}a^r\left(\frac{(1-a^{r+1})(1-a)}{1-a}\right)$
    $\displaystyle =\displaystyle \sum^n_{r=0}a^r(1-a^{r+1})$
    $\displaystyle \displaystyle \sum^n_{r=0}a^r-\sum^n_{r=0}a^{2r+1}$
    $\displaystyle =\frac{1-a^{n+1}}{1-a}-\frac{a-a^{2n+2}}{1-a^2}$

    $\displaystyle S_n=\frac{1-a^{n+1}}{(1-a)^2}-\frac{a-a^{2n+2}}{(1-a^2)(1-a)}$
    Which isn't the given one and I can't see where I've gone wrong
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    4
    Quote Originally Posted by arze View Post
    If $\displaystyle S_n=\displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)$, $\displaystyle (|a|\neq 1)$ by considering $\displaystyle (1-a)S_n$, show that
    $\displaystyle S_n=\frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
    State the set of values of a for which $\displaystyle S_n$approaches a limit as $\displaystyle n \rightarrow\infty$ and find the sum to infinity of the series for these values.

    Attempt:
    $\displaystyle \displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)$
    $\displaystyle =\displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^n_{r=0}a^r\right)$
    $\displaystyle =\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^r}{1-a}\right)$
    $\displaystyle =\displaystyle \sum^n_{r=0}\frac{a^r-a^{2r}}{1-a}$
    Am I correct so far? and how to continue because I don't see a way of summing that?
    Thanks!
    Dear arze,

    When n=1,

    $\displaystyle S_1=\displaystyle\sum_{r=0}^{1}{a^r(1+a+a^2+.....+ a^r)}=a^0+a(1+a)$

    But if you consider the expression, $\displaystyle \frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$......

    When n=1,

    $\displaystyle \frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$

    $\displaystyle =\frac{1-a^{4}}{(1-a)^3}-\frac{a^{2}(1+a^{2})}{(1-a)^2}$

    $\displaystyle =\frac{1+a^2}{1-a^2}\left(\frac{1-a^2}{1-a}-a^2\right)$

    $\displaystyle =\frac{1+a^2}{1-a^2}\left(1+a-a^2\right)$

    Therefore, $\displaystyle S_1\neq\frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
    Last edited by Sudharaka; Aug 17th 2010 at 02:23 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    Yea, that's not the correct identity. The correct identity is

    [tex]\displaystyle \displaystyle S_n = \frac{(1-a^{n+1})(1-a^{n+2})}{(1-a)^2(1+a)}[/Math]
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    Quote Originally Posted by Vlasev View Post
    Yea, that's not the correct identity. The correct identity is

    [tex]\displaystyle \displaystyle S_n = \frac{(1-a^{n+1})(1-a^{n+2})}{(1-a)^2(1+a)}[/Math]
    But that doesn't give me the result I'm supposed to get either
    $\displaystyle S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    Quote Originally Posted by arze View Post
    But that doesn't give me the result I'm supposed to get either
    $\displaystyle S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}$
    Because what you are asked to get is (in this form) false! Unless you typed something incorrectly in your question.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    hmmm, i checked the book again and i did not type anything wrong. Well, thank you very much!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    $\displaystyle \displaystyle\ S_n=\sum_{r=0}^na^r\left(1+a+a^2+....+a^r\right)$

    $\displaystyle \displaystyle\ aS_n=\sum_{r=0}^na^{r+1}\left(1+a+a^2+....+a^r\rig ht)$

    $\displaystyle \displaystyle\ (1-a)S_n=\sum_{r=0}^n\left(a^r+a^{r+1}+a^{r+2}+....+a ^{2r}\right)-\sum_{r=0}^n\left(a^{r+1}+a^{r+2}+....+a^{2r+1}\ri ght)$

    $\displaystyle \displaystyle\ =\sum_{r=0}^na^r-\sum_{r=0}^na^{2r+1}$

    The first sum has $\displaystyle n+1$ terms, the first term is 1 and the common ratio is $\displaystyle a$.
    The second sum has $\displaystyle (2n+2)/2 =n+1$ terms, the first term is $\displaystyle a$ and the common ratio is $\displaystyle a^2$.


    $\displaystyle \displaystyle\ (1-a)S_n=\frac{1-a^{n+1}}{1-a}-a\left(\frac{1-a^{2(n+1)}}{1-a^2}\right)$

    $\displaystyle \displaystyle\ S_n=\frac{1-a^{n+1}}{(1-a)^2}-a\left(\frac{1-a^{2n+2}}{(1-a)^2(1+a)}\right)$

    $\displaystyle \displaystyle\huge\ =\frac{(1+a)\left(1-a^{n+1}\right)-a\left(1-a^{2n+2}\right)}{(1-a)^2(1+a)}$

    $\displaystyle \displaystyle\huge\ =\frac{(1+a)\left(1-a^{n+1}\right)-a\left(1-a^{n+1}\right)\left(1+a^{n+1}\right)}{(1-a)^2(1+a)}$

    $\displaystyle \displaystyle\huge\ =\frac{\left(1-a^{n+1}\right)\left[1+a-a\left(1+a^{n+1}\right)\right]}{(1-a)^2(1+a)}$

    $\displaystyle \displaystyle\huge\ =\frac{\left(1-a^{n+2}\right)\left(1-a^{n+1}\right)}{(1-a)^2(1+a)}$

    so I'd have to agree with Vlasev.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 3rd 2011, 01:12 AM
  2. Replies: 3
    Last Post: Sep 29th 2010, 06:11 AM
  3. Replies: 0
    Last Post: Jan 26th 2010, 08:06 AM
  4. Replies: 1
    Last Post: May 5th 2008, 09:44 PM
  5. Replies: 11
    Last Post: Apr 1st 2008, 12:06 PM

/mathhelpforum @mathhelpforum