Results 1 to 4 of 4

Math Help - Sum to Infinity of a Geometric Series

  1. #1
    Member
    Joined
    Jul 2011
    Posts
    140

    Sum to Infinity of a Geometric Series

    Q.: The numbers 1/ t, 1/ (t - 1), 1/ (t + 2) are the first, second and third terms of a geometric sequence.
    Find (i) the value of t, (ii) the sum to infinity of the series.

    Eq.: S = a/ 1 - r

    Ans.: (i) t = 1/ 4, (ii) S = 6

    Attempt:
    (i) U2/ U1 = U3/ U2
    (1/ (t - 1))/ (1/ t) = (1/ (t + 2))/ (1/ (t - 1))
    (1/ (t - 1)) (1/ (t - 1)) = (1/ t) (1/ (t + 2))
    (1/ (t^2 - 2t + 1)) = (1/ (t^2 + 2t))
    (1/ (t^2 - 2t + 1 - t^2 - 2t))
    1/ (-4t + 1) this is where I lose the flow, slightly. Can someone help me confirm if my next move is permissible?

    1/ (4t + 1)
    1 = 4t
    t = 1/ 4

    (ii) S = a/ 1 - r
    a = 1/ (1/ 4) = 4, i.e. 1/ t
    r = U2/ U1 or U3/ U2 = (1/ ((1/ 4) - 1))/ 4 = (1/ (-3/ 4))/ 4 = (-4/ 3)/ 4 =
    (-4/ 3)(1/ 4) = -4/ 12 = -1/ 3
    4/ (1 - (-1/ 3)) This is where I think I may have gone wrong.
    4/ (1 + (1/ 3)) I think I need to have 1 - (1/ 3) here but I cant see how to bring that about.

    4/ (4/ 3)
    4 (3/ 4)
    12/ 4 = 3 The answer here should be 6 but I can only get 3.

    If I can get 1 - (1/ 3), above, Ill have (4/ (2/ 3)), and that gives 6 but as I have shown, the question does not work out that way for me. Have I missed something? Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426

    Re: Sum to Infinity of a Geometric Series

    I disagree that the sum is 6.

    \frac{1}{t} , \frac{1}{t-1} , \frac{1}{t+2} , ...

    common ratio of the first two terms ... r = \frac{t}{t-1}

    common ratio of the 2nd and 3rd terms ... r = \frac{t-1}{t+2}

    \frac{t}{t-1} = \frac{t-1}{t+2}

    (t-1)^2 = t(t+2)

    t^2 - 2t + 1 = t^2 + 2t

    1 = 4t

    t = \frac{1}{4}

    r = -\frac{1}{3}

    infinite sum ...

    \frac{4}{1 - \left(-\frac{1}{3}\right)} = 3
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Sum to Infinity of a Geometric Series

    t=\frac{1}{4} is correct, so the geometric sequence becomes: 4,\frac{-4}{3}, \frac{4}{9},...
    So r=\frac{-1}{3} and because \left|\frac{-1}{3}\right|<1 we can use: S=\frac{a}{1-r}=\frac{4}{1+\frac{1}{3}}=3

    So you're correct!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2011
    Posts
    140

    Re: Sum to Infinity of a Geometric Series

    Wow! I'm really surprised that they got that one wrong, in the book. Thanks for all the help, guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sum to Infinity of a Geometric Series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 19th 2011, 03:59 PM
  2. Replies: 5
    Last Post: August 11th 2011, 02:44 PM
  3. Sum to Infinity of a Geometric Series
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 7th 2011, 02:10 PM
  4. Sum to infinity of unusual geometric series
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: March 30th 2011, 08:03 PM
  5. The sum (to infinity) of a geometric series.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 25th 2009, 06:41 AM

Search Tags


/mathhelpforum @mathhelpforum