1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Binomial expansions

  1. Aug 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Find, in the simplest form, the coefficient of x^n in the binomial expansion of (1-x)^(-6).

    2. Relevant equations



    3. The attempt at a solution

    i am not sure how to go about with this.
     
  2. jcsd
  3. Aug 17, 2010 #2
    are u sure you have to find a cofficient contain x^n?

    because u should have a specific value for n, so that you could find the cofficient infront of it, or you should at least have which term you are looking for.
     
  4. Aug 18, 2010 #3
    Yes, that's the question. Maybe it's asking for the coefficient of for any term in the expression.
     
  5. Aug 18, 2010 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since the exponent, -6, is not a positive integer, you will need to use the generalized binomial series:
    [tex](a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}[/tex]

    Here, of course, a= 1, b= -x, and m= -6 so this is

    [tex](1- x)^{-6}= \sum_{k=0}^\infty \frac{-6(-7)\cdot\cdot\cdot(-5-k)}{k!}(-1)^kx^{-6-k}[/tex]

    You, apparently, are asked for the coefficient when -6-k= n or when k= -6-n.
     
  6. Aug 18, 2010 #5
    thanks

    Is this answer the most simplified?

    [tex]\frac{-6(-7)...(-11-n)}{(-6-n)!}[/tex]

    the general formula for binomial series for (a+b)^n is different when n is a positive integer and when n is a fractional or negative value?

    [tex](a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}[/tex]

    Does it matter if the powers(k and n-k) for a and b is swapped since a+b is commutative?

    This is the continuation of the question:

    Hence, find the coefficient of x^6 and x^7 in (1+2x+3x^2+4x^3+5x^4+6x^5+7x^6)^3
     
    Last edited: Aug 18, 2010
  7. Aug 19, 2010 #6
    any further hints on this question?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Binomial expansions
  1. Binomial expansion (Replies: 3)

  2. Binomial expansion (Replies: 1)

  3. Binomial Expansion (Replies: 6)

  4. Binomial expansion (Replies: 14)

  5. Binomial expansion (Replies: 2)

Loading...