Results 1 to 6 of 6

Math Help - summation

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    21

    summation

    If ak = 1/k-1/(k+1), then find out the summation of a2 to a100?
    (Here k, 2 & 100 are suffixes)

    a2=3
    a3=2
    a4=5/3 and so on i dont see any pattern how to find the summation ,is there any direct formula
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by arunachalam.s View Post
    If ak = 1/k-1/(k+1), then find out the summation of a2 to a100?
    (Here k, 2 & 100 are suffixes)

    a2=3
    a3=2
    a4=5/3 and so on i dont see any pattern how to find the summation ,is there any direct formula
    There is a pattern...

    \sum_{k=2}^{100}\frac{1}{k}-\frac{1}{k+1}=\frac{1}{2}{\color{red}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{100}}-\frac{1}{101}

    Everything in red cancels out and you're left with \sum_{k=2}^{100}\frac{1}{k}-\frac{1}{k+1}=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    21
    1/k-1/(k+1)=k+1/k-1 how did u get 1/k-1/k+1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by arunachalam.s View Post
    1/k-1/(k+1)=k+1/k-1 how did u get 1/k-1/k+1
    The way you posted your question could be interpreted in many ways...

    \frac{1}{k}-\frac{1}{k+1} or \frac{1}{k-\frac{1}{k+1}} or \frac{\frac{1}{k-1}}{k+1}.

    To make it easier for us to answer your future questions, please learn some basic \text{LaTeX}. See here for help.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Pim
    Pim is offline
    Member
    Joined
    Dec 2008
    From
    The Netherlands
    Posts
    91
    Order of operations.

    First divide, then subtract.

    Also, for the sanity of all of us, use LaTex, or at least use brackets.

    As I now see it, you could mean either one of these two:

    1/k-1/(k+1) =

    <br />
\frac{1}{k}-\frac{1}{k+1}<br />

    OR

    (1/(k-1))/(k+1) =

     <br />
\frac{\frac{1}{k-1}}{k+1} = \frac{k+1}{k-1}<br />
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2009
    Posts
    21
    i cant understand

    1/(k-1)/(k+1) lets put k=2 now this yields 1/(2-1)/(2+1)=1/3 which is not equal to (2+1)/(2-1)=3 i still cant understand how both the fractions r equal now if we take 1/k-1/(k+1) = 1/2-1/(3)=(3-2)/6 which is equal t0 1/6 now this both r not equal how u say all fractions are equal ..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to do a Summation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 14th 2011, 11:08 AM
  2. Summation
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: May 7th 2011, 06:23 AM
  3. summation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 2nd 2009, 08:17 AM
  4. Summation Help
    Posted in the Algebra Forum
    Replies: 9
    Last Post: January 31st 2009, 07:47 PM
  5. summation
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 8th 2008, 04:19 PM

Search Tags


/mathhelpforum @mathhelpforum