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Math Help - Generating function for a sequence

  1. #1
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    Generating function for a sequence

    The question is as follows:
    It is known that the generating function for the sequence , {1,8,27,64,...} is
    [x(x^2 + 4x +1)]/(1-x)^4
    Use the result to show summation from r=1 to n of r^3 = [n^2(n^2+1)]/4

    I know i have to find the sequence represented by 1,8,27,64, .. and once i get the sequence i have to manipulate it until i reach r^3 which will be given by the expression.
    My problem is that i cannot find the sequence for 1,8,27,64,...
    any help please???
    Thank buches!
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  2. #2
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    Hello, mehn!

    Please check the wording of the problem.
    None of it makes sense.


    It is known that the generating function for the sequence , {1, 8, 27, 64, ...}

    . . is: .[x(x^2 + 4x +1)]/(1-x)^4 . What?


    Use the result to show summation from r = 1 to n of r^3

    . .is: .[n^2(n^2+1)]/4 . Not true

    The generating function is: .f(r) = r^3
    . . The terms are consecutive cubes.

    The sum of the first n cubes is: .[n^2(n+1)^2]/4

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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, mehn!

    Please check the wording of the problem.
    None of it makes sense.



    The generating function is: .f(r) = r^3
    . . The terms are consecutive cubes.

    The sum of the first n cubes is: .[n^2(n+1)^2]/4

    Please read Generating Functions. The OP's generating function for cubes is correct.
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  4. #4
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    Opalg's Avatar
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    Quote Originally Posted by mehn View Post
    The question is as follows:
    It is known that the generating function for the sequence , {1,8,27,64,...} is
    [x(x^2 + 4x +1)]/(1-x)^4
    Use the result to show summation from r=1 to n of r^3 = [n^2(n^2+1)]/4
    This is a strange problem (to me, at any rate), but it actually works!

    The binomial series for (1-x)^{-4} is 1 + 4x + 10x^2 + 20x^3 + ... . When you multiply this by x(x^2 + 4x +1), you get x + 8x^2 + 27x^3 + 64x^4 + ... . The coefficients magically appear as the cubes of the integers!

    In this problem, you are asked to find the sum of the first n coefficients. One way to do that is to multiply the given generating function by (1-x)^{-1} and looking for the coefficient of x^n in that product.

    The coefficient of x^n in (1-x)^{-5} is . So the coefficient of x^n in is



    which eventually simplifies to [n^2(n+1)^2]/4.

    Not the easiest way to get that formula, but an interesting piece of mathematics.

    Edit. Thanks to alexmahone for that useful link. I hadn't seen his comment when writing mine.
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