Results 1 to 3 of 3

Math Help - Sigma solving

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    12

    Sigma solving

    How do i simplify this in terms of n?


    <br />
\sum_{i=0}^{n-1}{2^{i} (n-i)} <br />

    Cheers!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Quote Originally Posted by ukrobo View Post
    How do i simplify this in terms of n?


    <br />
\sum_{i=0}^{n-1}{2^{i} (n-i)} <br />

    Cheers!
    \sum_{i=0}^{n-1}{2^{i} (n-i)}=\frac{1}{2}\sum_{i=1}^{n}2^{i}+\frac{n}{2}\sum  _{i=1}^{n}2^{i}-\frac{1}{2}\sum_{i=1}^{n}i2^{i}

    Try using the partial sum for a geometric series formula.

    S_{n}=a_{n}\left(\frac{1-r^{n}}{1-r}\right), where

    r is the common ratio and a is the first term.

    So, from the first one we have \frac{1-2^{n}}{1-2}=2^{n}-1

    The second one is the same only multiplied by n. Giving n(2^{n}-1)

    The third only can be found by differentiating the partial sum formula.

    \sum_{i=1}^{n}x^{i}=\frac{x(1-x^{n})}{1-x}

    \frac{d}{dx}\left[\sum_{i=1}^{n}x^{i}\right]=\sum_{i=1}^{n}ix^{i-1}=\frac{(nx-n-1)x^{n}+1}{(x-1)^{2}}

    \sum_{i=1}^{n}ix^{i}=\frac{x((nx-x-1)x^{n}+1)}{(x-1)^{2}}

    Sub in x=2 and we get 2(2^{n}(n-1)+1).

    Don't forget to divide by 2 because of the 1/2.

    \frac{1}{2}\sum_{i=1}^{n}i2^{i}=(n-1)2^{n}+1

    Add this with the other two sums and we get:

    2^{n}-1+n(2^{n}-1)-((n-1)2^{n}+1)=\boxed{2^{n+1}-n-2}

    Just one way to go about it. Remember, you can play around with the geometric series formulas by differentiating, integrating, and so on, and get away with a lot.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,347
    Thanks
    30
    Quote Originally Posted by ukrobo View Post
    How do i simplify this in terms of n?


    <br />
\sum_{i=0}^{n-1}{2^{i} (n-i)} <br />

    Cheers!
    Here's another way. Let

     <br />
S_n = \sum_{i=0}^{n-1} 2^i(n-i) \;\;\;(1)<br />

    multiplying by 2 gives

     <br />
2S_n = \sum_{i=0}^{n-1} 2^{i+1}(n-i).<br />

    Now shift

     <br />
2S_n = \sum_{i=1}^{n} 2^i(n-i+1)\;\;\;(2).<br />

    Now substract (1) from (2) so

     <br />
S_n = \sum_{i=1}^{n} 2^i(n-i+1) - \sum_{i=0}^{n-1} 2^i(n-i)<br />
     <br />
= \sum_{i=1}^n 2^i - n = 2^{n+1} - n - 2.<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 4th 2011, 08:39 AM
  2. Show intersection of sigma-algebras is again a sigma-algebra
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 20th 2010, 07:21 AM
  3. how to determine sigma and sgn(sigma) for matrices
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 16th 2010, 11:12 AM
  4. help with solving equations using sigma notation
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 13th 2009, 07:18 AM
  5. how to get sigma
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: September 25th 2009, 07:07 PM

Search Tags


/mathhelpforum @mathhelpforum