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Binomial theorem problem on the terms of an expansion

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Find an approximation of (0.99)5 using the first three terms of its expansion.

    2. The attempt at a solution
    To get to the binomial theorem I divided 0.99 into
    (0.99)5 = (1-0.01)5 = {1+(-0.01)}5
    Then,
    T1 = 5C0(1)5 = 1 x 1=1
    T2 = 5C1(1)5-1(-0.01)1 = 5x1x -0.01= (-0.05)
    T3 = 5C2(1)5-2(-0.01)2 = 10 x 1x 0.0001 = (0.001)

    Now do I add them?? But adding them doesn't get me to the answer. Please help. Thanks in advance.
     
  2. jcsd
  3. May 16, 2012 #2

    Ray Vickson

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    Why do you say "adding them doesn't get me to the answer"? Of course, it does not get you to the exact value of 0.99^5, but that is not the issue. Just adding the terms you have fulfills all the requirements of the problem.

    RGV
     
  4. May 17, 2012 #3
    OK thanks for the help! Actually I had some confusion about that operation.
    Um....I's thinking that will I get to the same answer if I had divided 0.99 into (0.9+0.09)?
     
  5. May 17, 2012 #4

    HallsofIvy

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    You would get another approximation, not necessarily the same answer. You would have to calculate the fifth, fourth, and third powers of .9 which is harder than the same powers of 1!
     
    Last edited: May 17, 2012
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