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Homework Help: Sum of Geometric Sequence

  1. Jun 4, 2005 #1
    I'm given the sequence t(n) = 3 (-1)^n (0.5)^n ; n >= 1

    It first asks for the sum of the terms t(1) + t(2) + ... + t(99) which is fine, but it follows up by asking the sum of t(1) + t(3) + t(5) + ... + t(99).

    Would i be using partial sums to solve this? If so, i don't know how to find the sum of (-0.5)^2n-1 for n = 1 to 99.

    Any help would be appreciated, thanks in advance.
     
  2. jcsd
  3. Jun 5, 2005 #2
    never mind...

    you know, sometimes i wonder about myself...
     
  4. Jun 16, 2005 #3
    Give you a hint...this is a geometric sequence
    where 3(-1)^n(0.5)^n = 3(-.5)^n
     
  5. Jun 16, 2005 #4
    as quentinchin said, this is a geometric sequence
    t(n) = 3(-1)^n(0.5)^n = 3(-0.5)^n

    now,
    t(1) + t(3) + t(5) + ...... + t(99)
    = 3 [(-0.5)^1 + (-0.5)^3 + (-0.5)^5 + ...... + (-0.5)^99 ]

    now the series inside the brackets [] has first term, a = (-0.5)^1 = -0.5 and common ratio r = (-0.5)^2 = 0.25

    therefore,
    3 [(-0.5)^1 + (-0.5)^3 + (-0.5)^5 + ...... + (-0.5)^99 ]
    = 3 [a/(1-r)]
    = 3 [-0.5/(1-0.25)]
    = 3 [-0.5/0.75]
    = -2
     
  6. Jun 17, 2005 #5
    i made a little mistake in my last reply.

    t(1) + t(3) + t(5) + ...... + t(99)
    = 3 [(-0.5)^1 + (-0.5)^3 + (-0.5)^5 + ...... + (-0.5)^99 ]

    now the series inside the brackets [] has first term, a = (-0.5)^1 = -0.5, common ratio r = (-0.5)^2 = 0.25 and number of terms, n = 50

    therefore,
    3 [(-0.5)^1 + (-0.5)^3 + (-0.5)^5 + ...... + (-0.5)^99 ]
    = 3 [a(1 - r^n)/(1-r)]
    = 3 [-0.5(1-(-0.5)^50)/(1-0.25)]

    but (-0.5)^50 is so close to zero that we can ignore that. therefore,

    3 [-0.5{1-(-0.5)^50}/(1-0.25)]
    = 3[-0.5(1-0)/(1-0.25)]
    = 3 [-0.5/0.75]
    = -2
     
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