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Homework Help: Help! Binomial Theorem

  1. Jul 2, 2005 #1
    Find the term containing [tex]x^{20}[/tex] in [tex](2x - x^4)^{14}[/tex].

    I went [tex]t_{k+1}= _{14}C_{k}(2x)^{14-k}(-x^{4})^{k}[/tex]

    [tex]= 2x^{14-k}(-x^{4k})[/tex]

    First of all, am I on the right track? If so what exactly do I do from there?
     
  2. jcsd
  3. Jul 2, 2005 #2

    AKG

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    Something is definitely wrong. First of all, your binomial coefficient has disappeard on the second line. Secondly, you seem to pull the 2 out of (2x)14-k in an improper way. You also can't bring the k inside the bracket like that, since it's not ((-x)4)k, it's (-x[sup4[/sup])k. Do you see the difference? The exponent 4 applies only to the x, not to (-x), but the exponent k applies to the whole (-x4). You should end up with:

    [tex]t_{k+1} = {{14}\choose k}(-2)^{14-k}x^{14 + 3k}[/tex]

    See if you can figure out why the above is right. Now, to find the term containing x20, find the value(s) of k that satisfy 14 + 3k = 20. The only value for k is 2. So plug in 2 for k, you'll get:

    (91)(212)x20
     
  4. Jul 2, 2005 #3
    Okay I think I get your point about the [tex]-x^4 [/tex] thing. But how did you get the 2x to turn into -2 and how did you get the 14+3k exponent?
     
  5. Jul 2, 2005 #4

    lurflurf

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    You want the x^20 term so if the exponent of x in the kth term is
    4k+(14-k)=20 what is k?
     
  6. Jul 3, 2005 #5
    +3k... right...thanks I see it now!
     
  7. Jul 3, 2005 #6
    Take (x^14) common and find the third term in the expansion of (2-(x^3))^14 i.e.
    14C2*(2^12)*(-1)^2.
     
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