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Math Help - Geometric Progressions

  1. #1
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    Geometric Progressions

    Hi all.

    I'm trying to prove that

    ( 1 + x + x^2 + x^3 + ... + x^2k) ( 1 - x + x^2 - x^3 + ... + x^2k)

    = 1 + x^2 + x^4 + ... + x^4k, where k is a positive integer and x is not equal to -1, 1.

    This question was posted in the section on Geometric Progressions.

    I don't know how to begin with this... is x^2k the last term? What does it mean, is it a power of 2?? For example x raised to some power of 2??

    Do I use the summation formula S_n = a ( 1 - r^n) / (1 -r)??

    Help will be appreciated.

    Thanks.
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  2. #2
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    Quote Originally Posted by pollardrho06 View Post
    Hi all.

    I'm trying to prove that

    ( 1 + x + x^2 + x^3 + ... + x^2k) ( 1 - x + x^2 - x^3 + ... + x^2k)

    = 1 + x^2 + x^4 + ... + x^4k, where k is a positive integer and x is not equal to -1, 1.

    This question was posted in the section on Geometric Progressions.

    I don't know how to begin with this... is x^2k the last term? What does it mean, is it a power of 2?? For example x raised to some power of 2??

    Do I use the summation formula S_n = a ( 1 - r^n) / (1 -r)??

    Help will be appreciated.

    Thanks.
    Yes, use the summation formula...you are summing n=2k+1 terms

    1+x+x^2+.....x^{2k}

    the nth term is ar^{n-1}\ \Rightarrow\ for\ x^{2k}\ \Rightarrow\ n=2k+1

    a=1,\ r=x

    S_n=\frac{1-x^{2k+1}}{1-x}


    1-x+x^2-x^3+...+x^{2k}

    r=-x

    S_n=\frac{1-(-x)^{2k+1}}{1-(-x)}

    2k+1 is odd as 2k is even, hence

    (-x)^{2k+1}=-\left(x^{2k+1}\right)

    S_n=\frac{1+x^{2k+1}}{1+x}

    The product of these sums is

    \left(\frac{1-x^{2k+1}}{1-x}\right) \frac{1+x^{2k+1}}{1+x}=\frac{1-x^{4k+2}}{1-x^2}


    1+x^2+x^4+....+x^{4k}=1+(x)^2+\left(x^2\right)^2+.  ...+\left(x^{2k}\right)^2

    S_n=\frac{1-\left(x^2\right)^{2k+1}}{1-x^2}=\frac{1-x^{4k+2}}{1-x^2}
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    Yes, use the summation formula...you are summing n=2k+1 terms

    1+x+x^2+.....x^{2k}

    the nth term is ar^{n-1}\ \Rightarrow\ for\ x^{2k}\ \Rightarrow\ n=2k+1

    a=1,\ r=x

    S_n=\frac{1-x^{2k+1}}{1-x}


    1-x+x^2-x^3+...+x^{2k}

    r=-x

    S_n=\frac{1-(-x)^{2k+1}}{1-(-x)}

    2k+1 is odd as 2k is even, hence

    (-x)^{2k+1}=-\left(x^{2k+1}\right)

    S_n=\frac{1+x^{2k+1}}{1+x}

    The product of these sums is

    \left(\frac{1-x^{2k+1}}{1-x}\right) \frac{1+x^{2k+1}}{1+x}=\frac{1-x^{4k+2}}{1-x^2}


    1+x^2+x^4+....+x^{4k}=1+(x)^2+\left(x^2\right)^2+.  ...+\left(x^{2k}\right)^2

    S_n=\frac{1-\left(x^2\right)^{2k+1}}{1-x^2}=\frac{1-x^{4k+2}}{1-x^2}
    Gr8, thanks!!
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