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Evaluate finite sum

  1. Nov 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Find [itex]\sum\limits_{k=0}^{n}k^2{n\choose k}(\frac{1}{3})^k(\frac{2}{3})^{n-k}[/itex]

    2. Relevant equations
    -Binomial theorem

    3. The attempt at a solution
    I am using the binomial coefficient identity [itex]{n\choose k}=\frac{n}{k}{{n-1}\choose {k-1}}[/itex]:

    [tex]\sum\limits_{k=0}^{n}k^2{n\choose k}(\frac{1}{3})^k(\frac{2}{3})^{n-k}=\sum\limits_{k=1}^{n-1}\frac{n}{k}{{n-1}\choose {k-1}}(k-1)^2(\frac{1}{3})^{k-1}(\frac{2}{3})^{n-k+2}[/tex]

    What am I doing wrong here (sums are not equal)?
     
  2. jcsd
  3. Nov 26, 2015 #2
    How do you compute the exponent of 2/3 ?
     
  4. Nov 26, 2015 #3
    What do you mean? In the sum, [itex]k[/itex] is increased by [itex]1[/itex] and [itex]n[/itex] is decreased by [itex]1[/itex], so in the function [itex]k[/itex] is decreased and [itex]n[/itex] is increased.
     
  5. Nov 26, 2015 #4
    I think first we shall note the domain of validity of the binomial formula.

    This imply we shall write a term outside the sum.

    Are the k modified in the sum out side the binomial formula ?

    I suppose the aim of using that formula two times is to get rid of the k squared.
     
  6. Nov 26, 2015 #5
    In general, you have ##(x+y)^n = \sum_{k = 0}^n \binom{n}{k} x^k y^{n-k} ##

    Take the derivative with respect to ##x## on each sign of the equation and multiply by ##x##.
    You get ##nx(x+y)^{n-1} = \sum_{k = 0}^n k \binom{n}{k} x^k y^{n-k} ##.

    Now do it again and set ##x## and ##y## to one third and two thirds.
     
  7. Nov 26, 2015 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Apparently you are trying to do this problem without knowing what [itex]\sum_{k= 0}^n [/itex] means.
    n does NOT "decreas", n is fixed, the maximum value of k.
     
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