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How can I find the coefficient of x^12

  1. Nov 23, 2004 #1
    How can I find the coefficient of x^12 in:
    [tex] (2+x)^{14}(1+ \frac{2}{x})^{14} [/tex]
    I did it like that:
    [tex] (2+x)^{14}(1+ \frac{2}{x})^{14} [/tex]
    Can be written as:
    [tex] (4+ \frac{4+x^2}{x})^{14} [/tex]
    Now the term [tex]T_{r+1}[/tex] can be represented as:
    [tex] 14C_r (4)^{14-r}(\frac{4+x^2}{x})^r[/tex] (How can I write 14Cr?)
    Dont know where to go from here?

    Thanks in advance for any help. :smile:
     
    Last edited: Nov 23, 2004
  2. jcsd
  3. Nov 23, 2004 #2

    matt grime

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    There's more than one way to skin a cat:

    how can you obtain x^12 from the two original brackets? You need x^14 from one and x^{-2} from the other and... add 'em up.

    for the tex for binomial coeffs click on:

    [tex]\binom{n}{r}[/tex]
     
  4. Nov 23, 2004 #3
    Thanks alot for your reply Matt. I got what you said but I'm still curious to find out how can I find the answer to it by using the method I stated above? I have done many similar questions and I know I have to find the value of r for which the power of x is 12. Is this possible?
     
    Last edited: Nov 23, 2004
  5. Nov 23, 2004 #4

    matt grime

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    of course it is- just not very easy, you just need, for each r to find the positive integer s where 0<=s<=r, if it exists, such that 2s-r=12 and then add up, for each one where this happens,

    [tex]\binom{14}{r}\binom{r}{s}4^{14-r}4^{r-s}[/tex]


    hardly elegant but then you made the problem much harder with your choice of "simplification"
     
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