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Binomial Expansions.

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations
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    3. The attempt at a solution
    After using this formula I got 1-3x + 6x2 for (1-x)-3 and 1+x - 1/8 ∙ 4x2 for (1+2x)1/2 for the second part of the question I'd assume that you're supposed to multiply the equations? I think this done by timing the whole of second bracket by first number (1) then second number (-3x) and then third number (6x2.from the second bracket.
     
  2. jcsd
  3. Feb 9, 2016 #2

    blue_leaf77

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    Check again the sign of the second term.
     
  4. Feb 9, 2016 #3
    I don't see where I'm going wrong here? (-3)(-4)(x)2(1/2) = 6x2
     
  5. Feb 9, 2016 #4

    blue_leaf77

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    The second term, not the third one.
     
  6. Feb 9, 2016 #5

    PeroK

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    No one says "timing". The word is "multiplying". But, yes, that's just the distributive law.
     
  7. Feb 9, 2016 #6
    Sorry, I'm not seeing the mistake? -3(x) = -3x correct?
     
  8. Feb 9, 2016 #7

    blue_leaf77

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    The formula for the binomial expansion uses a plus sign in front of ##x## to define the equation: ##(1+x)^n##. While in the question, the sign of ##x## is minus.
     
  9. Feb 9, 2016 #8
    Oh I was just looking at it as +(-3) which is same as just -3 so I can't use this formula?
     
  10. Feb 9, 2016 #9
    My mistake, would you mind checking to see if I did the first part correct?
     
  11. Feb 9, 2016 #10

    blue_leaf77

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    You can use that formula, but the sign of each ##x## appearing in the RHS (the infinite sum) must be adjusted accordingly. I will rewrite the formula using different variable notation for the sake of clarity
    $$
    (1+u)^{-3} = 1-3u+6u^2-10u^3+\ldots
    $$
    what if you change ##u\rightarrow -x## as you have in the problem?
     
  12. Feb 9, 2016 #11
    ooohh it all becomes negative? I see! So its 1-3x - 6x2?
     
  13. Feb 9, 2016 #12

    blue_leaf77

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    No, that's not how you go about this.
    Ok rather than an arrow sign, I will use an equality sign. Make a change ##u=-x## in that formula. Does it help this time?
     
  14. Feb 9, 2016 #13

    PeroK

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    It's not the binomial theorem that is the problem, it's your basic algebra, especially handling negative signs. I suggest you do a bit of revision of this.
     
  15. Feb 9, 2016 #14
    1-3-x + 6-x2 - 10-x3
     
  16. Feb 9, 2016 #15

    blue_leaf77

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    I know what you did there is simply replacing ##u## with ##-x## without giving heed to the rules of writing the accepted mathematical expression. What does -3-x in that expression mean?
     
  17. Feb 9, 2016 #16
    I'm not sure how it should be expressed. -3-x = -(x+3). Should it be 1 - (3+x)? (IGNORE)

    Its (-3)(-x) so it should be 3x because its -3 * -x. sorry
     
  18. Feb 9, 2016 #17

    blue_leaf77

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    Yes, it should have been ##+3x##.
     
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