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Binomial Expansion Question

  1. Dec 29, 2012 #1
  2. jcsd
  3. Dec 29, 2012 #2

    Dick

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  4. Dec 29, 2012 #3
    You got the correct answer somehow, but your working is flawed. Can you see where?
     
  5. Dec 29, 2012 #4

    Dick

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    Ooops. Yeah, I agree. I somehow just read through the bad part. -1/0 should have been a tip off.
     
  6. Dec 30, 2012 #5
    oh yes -1/0 is -∞ but I made it zero:frown:
    a/3 +1 = -1/2x
    putting x = 0
    a/3 +1 = -1/2(0)
    a/3 +1 = -1/0
    -1/0 is -∞
    a/3 +1 = -∞
    if a = -3 is the correct answer how to get this value :confused:
     
    Last edited: Dec 30, 2012
  7. Dec 30, 2012 #6
    The question states "the coefficient of the term in x is zero".

    What do you think this coefficient is?
     
  8. Dec 30, 2012 #7
    somebody told me this general formula
    [itex]T_{r+1} = \binom{n}{r}a^n b^r[/itex]
    will be used to find 'a' and the statement "the coefficient of the term in x is zero" means
    that [itex]\binom{n}{r}[/itex] is 0 and what I did previously is wrong.
    I have math exam tomorrow and this is the only question that I cannot solve
     
    Last edited: Dec 30, 2012
  9. Dec 30, 2012 #8

    Dick

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    What you did before is almost right. When you get to 1+(2ax/3)+2x=1+(2a/3+2)x the part you want to make 0 is just the coefficient of x, (2a/3+2). Ignore the 1, it doesn't have anything to do with x.
     
  10. Dec 30, 2012 #9
    @Dick
    can you please show me last two steps of how to solve it for a
     
  11. Dec 30, 2012 #10

    Dick

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    Ok, just for you. 2a/3+2=0, 2a/3=(-2) (subtract 2 from both sides), 2a=(-2)*3 (multiply both sides by 3), a=(-2)*3/2=(-3) (divide both sides by 2).
     
  12. Dec 30, 2012 #11
    Thanks
     
  13. Dec 30, 2012 #12

    Dick

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    You're welcome. Notice no 1/0 appears. If it does that's a pretty sure sign something is wrong.
     
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