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Math Help - Find the generating function for Ar and use it to evaluate Ar...

  1. #1
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    Find the generating function for Ar and use it to evaluate Ar...

    This deals with summation methods for generating functions...

    Let Ar = C(2,2) + C(3,2) + ... + C(r+2,2). Find the generating function for Ar and use it to evaluate Ar.

    b) Do the same for Br
    Let Br = C(n,n) + C(n+1,n) + ... + C(n+r,n) where n is a given positive integer.
    So for part b I need to find a generating function for Br and use it to evaluate Br.

    Any help please I'm not sure how to form a generating function, and I missed this day of class, and this is the only problem on these summation methods.
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  2. #2
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    Re: Find the generating function for Ar and use it to evaluate Ar...

    \left( \sum_{k = 2}^{\infty} C(k, 2)x^k \right)(1 + x + x^2 + x^3 +...)

    = [C(2,2)]x^2 + [C(3,2) + C(2,2)]x^3 + [C(4, 2) + C(3,2) + C(2,2)]x^4 + ...

    \text{so } \left( \sum_{k = 2}^{\infty} C(k, 2)x^k \right)\left( \frac{1}{1-x} \right) = \sum_{k = 2}^{\infty} A_{k-2}x^k.

    \text{Now }\sum_{k = 2}^{\infty} C(k, 2)x^k = x^2\sum_{k = 0}^{\infty} C(k+2, 2)x^k

    \text{and }C(k+2, 2)x^k = \frac{(k+2)!}{2! \ k!}x^k = \frac{1}{2} (k+2)(k+1)x^k = \frac{1}{2} \frac{d^2}{dx^2}(x^{k+2})

    \text{so }\sum_{k = 2}^{\infty} C(k, 2)x^k = x^2\sum_{k = 0}^{\infty} \frac{1}{2} \frac{d^2}{dx^2}(x^{k+2})

    = \frac{x^2}{2} \frac{d^2}{dx^2} \left(\sum_{k = 0}^{\infty} x^{k+2} \right)

    = \frac{x^2}{2} \frac{d^2}{dx^2} \left(x^2\sum_{k = 0}^{\infty} x^k \right)

    = \frac{x^2}{2} \frac{d^2}{dx^2} \left(\frac{x^2}{1-x} \right)

    = \frac{x^2}{2} \frac{d}{dx} \left(\frac{(2x)(1-x) - (x^2)(-1)}{(1-x)^2} \right)

    = \frac{x^2}{2} \frac{d}{dx} \left(\frac{2x-x^2}{(1-x)^2} \right)

    = \frac{x^2}{2} \left( \frac{(2-2x)(1-x)^2 - (2x-x^2)[2(1-x)(-1)]}{(1-x)^4} \right)

    = \frac{x^2}{2} \left( \frac{2 \ [ \ (1-x)^2 + (2x-x^2) \ ]}{(1-x)^3} \right)

    = \frac{x^2}{2} \left( \frac{2}{(1-x)^3} \right)

    = \frac{x^2}{(1-x)^3}.

    \text{Thus } \sum_{k = 2}^{\infty} A_{k-2}x^k = \left( \sum_{k = 2}^{\infty} C(k, 2)x^k \right)\left( \frac{1}{1-x} \right)

    = \left( \frac{x^2}{(1-x)^3} \right)\left( \frac{1}{1-x} \right) = \frac{x^2}{(1-x)^4}.

    \text{Since } \sum_{k = 2}^{\infty} A_{k-2}x^k = x^2 \sum_{k = 0}^{\infty} A_kx^k,

    \text{have } x^2 \sum_{k = 0}^{\infty} A_kx^k = \frac{x^2}{(1-x)^4}.

    \text{Therefore } \sum_{k = 0}^{\infty} A_kx^k = \frac{1}{(1-x)^4}.

    This the generating function for A_k.

    Now A_k is calcuable by finding the Taylor series for that function about x=0. Its derivatives are very easy to compute and very simple to write down in general. I'll leave that to you.
    Last edited by johnsomeone; October 17th 2012 at 02:30 AM.
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