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Cofficent ?

  1. Sep 15, 2004 #1
    What is the coeffient of x^34 when n=1000?

  2. jcsd
  3. Sep 15, 2004 #2
    what have you done so far?
  4. Sep 15, 2004 #3
    what ? so far ?
    i stay at one place only. no farther, very near!
    means i don no how to solve problem.
  5. Sep 15, 2004 #4


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    Do you know anything about the "binomial theorem" or "multinomial theorem". If not, why in the world are you doing a problem like this?
  6. Sep 15, 2004 #5
    Speaking of the multinomial theorem, does anyone know where i can get sample problems? I have a test next tuesday that is on the binomial and multinomial theorem, unfortunately my textbook has none.
  7. Sep 16, 2004 #6
    You may like to get your hands on Hall & Knight, if its possible for u to get it through some library maybe.
  8. Sep 16, 2004 #7
    Well, to present a "hands on" start on this, you have to look at all the ways you can come up with X^34, and the exponents on these algebraic terms total 1000. (We need a lot of 1s.) One way is (X^16)^2(X^2)^1(1^997). (Hopefully you understand what I mean.) Coefficient for this is then (1000!)/[2!1!997!].

    Then we proceed to look at (X^16)^1(X^7)^2(X^2)^2(1^997), this gives us the coefficient: (1000!)/[1!2!997!].

    So proceeding in this method to look at all cases, we then add up all the coefficients, and presumedly, that is the correct answer.

    A simpler question would be to find the coefficient for X^3 on (1+X+X^2)^4 .

    There is only two ways: (X)^1(X^2)^1(1^2), and (x)^3(1^1). So the coefficient is 4!/(2!1!1!) and 4!/(3!1!) = 12 + 4 = 16.

    Check out: http://encyclopedia.thefreedictionary.com/Trinomial theorem
    Last edited: Sep 16, 2004
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