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Binomial series

  1. Apr 27, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex] g(x) = \sum_{n=0}^\infty \binom{k}{n} x^n [/itex]

    [itex] g'(x) = k\sum_{n=0}^\infty \binom{k-1}{n} x^n [/itex]

    prove that (1+x)g'(x) = kg(x)

    3. The attempt at a solution

    [itex] k(1+x)\sum_{n=0}^\infty \binom{k-1}{n} x^n [/itex]


    [itex] k[\sum_{n=0}^\infty \binom{k-1}{n} x^n + x\sum_{n=0}^\infty \binom{k-1}{n} x^n] [/itex]

    [itex] k[\sum_{n=0}^\infty \binom{k-1}{n} x^n + \sum_{n=0}^\infty \binom{k-1}{n} x^{n+1}] [/itex]

    [itex] k[\sum_{n=0}^\infty \binom{k-1}{n} x^n + \sum_{n=1}^\infty \binom{k-1}{n-1} x^n] [/itex]

    This is where I am stuck. I want to be able to pull out the x^n and add [itex] \binom{k-1}{n} + \binom {k-1}{n-1} [/itex] because I already know when you add those it gives you [itex] \binom{k}{n}[/itex] and I would have my answer, but one series starts at one and the other 0 so I can't pull them out. Please help
  2. jcsd
  3. Apr 28, 2014 #2


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    Just separate out the first term:
    $$\sum_{n=0}^\infty \begin{pmatrix}k-1 \\ n\end{pmatrix}x^n = 1 + \sum_{n=1}^\infty \begin{pmatrix}k-1 \\ n\end{pmatrix}x^n$$
  4. Apr 28, 2014 #3
    Ahhh okay thank you! So:

    [itex] k[1 + \sum_{n=1}^\infty \binom{k-1}{n} x^n + \sum_{n=1}^\infty \binom{k-1}{n-1} x^n] [/itex]

    [itex] k[1 + \sum_{n=1}^\infty x^n ( \binom{k-1}{n} + \binom{k-1}{n-1})] [/itex]

    [itex] k[1 + \sum_{n=1}^\infty x^n \binom{k}{n}] [/itex]

    [itex] k\sum_{n=0}^\infty x^n \binom{k}{n} [/itex]
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