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Binomial Theorem - Determine n

  1. Feb 12, 2017 #1
    1. The problem statement, all variables and given/known data
    The sixth term of the expansion of (x-1/5)n is -1287/(3125)x8. Determine n.

    2. Relevant equations
    tk+1=nCkan-kbk

    3. The attempt at a solution
    tk+1=nCkan-kbk
    t5+1=nC5(x)n-5(-1/5)5
    This is where I'm stuck. Do I sub in -1287/(3125)x8 to = t6? If so what do I do from here?
    -1287/(3125)x8 = nC5(x)n-5(-1/5)5?
    Any help is greatly appreciated!
     
  2. jcsd
  3. Feb 12, 2017 #2

    fresh_42

    Staff: Mentor

    You already wrote ##(-\frac{1}{5})^5=-\frac{1}{3125}## and ##x^{n-5}=x^8##, what is ##n## then? ##\binom{n}{5}## should only serve as a control calculation.
     
  4. Feb 12, 2017 #3
    So all I need is the x8 and xn-5?
    Then bases are the same so - n-5 = 8 ----> n=3?
     
  5. Feb 12, 2017 #4

    fresh_42

    Staff: Mentor

    ##3-5=8## ?
     
  6. Feb 12, 2017 #5
    Whoops! Sorry I meant to put n = 13!
     
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