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Can someone help me understand the trinomial expansion?

  1. Dec 13, 2007 #1

    rock.freak667

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    Homework Helper

    Well according to what I've read

    [tex](a+b+c)^n=\sum_{i,j,k} \left(
    \begin{array}{c}
    n\\
    i,j,k
    \end{array}
    \right)a^i b^j c^k[/tex]



    [tex]
    \left(
    \begin{array}{c}
    n\\
    i,j,k
    \end{array}
    \right)
    =\frac{n!}{i!j!k!}[/tex]

    I understand the last equation but how would I find the values for i,j and k?

    for example if I have [itex](1+x+x^2)^8[/itex] how would I find the coefficient of x^3 without expanding the entire thing out?
     
  2. jcsd
  3. Dec 13, 2007 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    This:

    [tex]\sum_{i,j,k}[/tex]

    means to sum over all (i,j,k) from 0 to n such that i+j+k = n.
     
  4. Dec 13, 2007 #3
    In other words, find (i,j,k) such that (1^i)*(x^j)*(x^2)^k = x^3 and i+j+k = 8. Then just evaluate that particular multinomial coefficient.
     
  5. Dec 13, 2007 #4

    rock.freak667

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    so from:(1^i)*(x^j)*(x^2)^k = x^3
    j+2k=3 and i+j+k=8...but since I dont have a 3rd equation how would i find those specific values for i,j and k ?
     
  6. Dec 13, 2007 #5

    Ben Niehoff

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    Gold Member

    There are two distinct possibilities. You will have to find both trinomial coefficients and add them together.

    Hint: You know j and k are integers between 0 and 8. Just try stuff.
     
  7. Dec 13, 2007 #6

    rock.freak667

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    Well the only way j+2k=3 would be if j=1,k=1 and so i=6?

    then the coefficient of x^3 would be 56?
     
  8. Dec 13, 2007 #7

    Ben Niehoff

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    Remember that j and/or k can also be zero, so there is one additional possibility.

    You're very close, though. :)
     
  9. Dec 13, 2007 #8

    rock.freak667

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    Well after picking k=0 I got the coefficient to now be 112 and i think that is all i can guess j and k to be.
     
  10. Jun 10, 2011 #9
  11. Jun 11, 2011 #10

    HallsofIvy

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    Staff Emeritus
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    With a= 1, b= x, [itex]c= x^2[/itex], [itex]a^ib^jc^k[/itex] is equal to [itex]1^ix^j(x^2)^k= x^{j+2k}= x^8[/itex] so j+ 2k= 8 as said before. If k= 0, j= 8. If k= 1, j= 6. If k= 2, j= 4. If k= 3, j= 2. If k= 4, j= 0. The condition that i+ j+ k= 8 gives i= 0, 1, 2, 3, and 4 respectively.

    Therefore, the coefficient of [itex]x^8[/itex] is the sum of
    [tex]\begin{pmatrix}8 \\ 0, 8, 0\end{pmatrix}+ \begin{pmatrix}8 \\ 1, 6, 1\end{pmatrix}+ \begin{pmatrix}8 \\ 2, 4, 2\end{pmatrix}+ \begin{pmatrix}8 \\ 3, 2, 3\end{pmatrix}+ \begin{pmatrix}8 \\ 4, 0, 4\end{pmatrix}[/tex]
     
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