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Math Help - Sum of Geometric series........

  1. #1
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    Sum of Geometric series........

    Sum to n groups the following series

    (x+y) + (x^2+xy+y^2) + {x^3+(x^2)*y+x(y^2)+y^3}

    Need help fast..............Thanks a lot in advance.
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  2. #2
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    Sum to n groups the following series

    (x+y) + (x^2+xy+y^2) + (x^3+(x^2)*y+x(y^2)+y^3)

    Need help fast..............Thanks a lot in advance.[/QUOTE]

    Sorry..... I forgot to give brackets in the 3rd group........
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  3. #3
    Junior Member RHandford's Avatar
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    And what have you done or tried so far?
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  4. #4
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    Quote Originally Posted by Arka View Post
    Sum to n groups the following series

    (x+y) + (x^2+xy+y^2) + (x^3+(x^2)*y+x(y^2)+y^3)

    Need help fast..............Thanks a lot in advance.

    sure you didn't mean ...

    (x+y) + (x^2 + 2xy + y^2) + (x^3 + 3x^2y + 3xy^2 + y^3) + ...

    ???
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  5. #5
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    No I did not mean that.........

    I tried till

    (x+y) + (x(x+y) + y^2) + (x(x^2+xy+y^2) + y^3) + .......

    Now, from here if I get the sum of the nth group , I can easily find the sum to n groups.............but that's what I'm having trouble finding.......
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  6. #6
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    Quote Originally Posted by skeeter View Post
    sure you didn't mean ...
    (x+y) + (x^2 + 2xy + y^2) + (x^3 + 3x^2y + 3xy^2 + y^3) + ... ???
    Quote Originally Posted by Arka View Post
    No I did not mean that.........
    Then that is not a geometric series.
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  7. #7
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    NO.....It's not......but see....

    (x+y) + (x(x+y) + y^2) + (x^2(x+y) + y^2(x+y)) +......

    I definitely see a pattern up there... and I belive it is possible to find the sum of the nth group and then the sum to n groups.........but I can't figure out how to do it........

    And the chapter in which this sum is given is called "Geometrical Progression".......so I believe that even if the series as a whole is not in G.P. some part of it must be in G.P.
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  8. #8
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    x^{4}-y^{4} = (x-y)(x+y)(x^{2}+y^{2})

    therefore, Series reduces to,

    (x^{2}-y^{2})/(x-y) + (x^{3}-y^{3})/(x-y) + (x^{4}-y^{4})/(x-y)+....

    Simplify further, you'll obtain something resembling a geometric series

    Obviously this is under the assumption that x and y are not equal.
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  9. #9
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    How do I simplify further.......?
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  10. #10
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    See the common denominator,

    u have, (1/(x-y))((x^{2}+x^{3}+......)-(y^{2}+y^{3}+......)) upto n terms

    so there are two geometric progressions in x and y. Now sum each upto n terms
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  11. #11
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    Hello, Arka!

    Sum to \,n groups the following series:

    . . S_n \;=\;(x+y) + (x^2+xy+y^2) + (x^3 + x^2y+xy^2+y^3) + \hdots

    Multiply each group by \frac{x-y}{x-y}

    . . \begin{array}{ccccccc}<br />
\dfrac{x+y}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^2-y^2}{x-y} \\ \\[-3mm]<br />
\dfrac{x^2+xy+y^2}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^3-y^3}{x-y} \\ \\[-3mm]<br />
\dfrac{x^3+x^2y+xy^2+y^3}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^4 - y^4}{x-y} \\ \\[-3mm]<br />
\vdots && \vdots \\<br />
\dfrac{x^n + x^{n-1}y + \hdots + xu^{n-1} + y^n}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^{n+1}-y^{n+1}}{x-y} \end{array}



    \text{Hence: }\;S_n \;=\;\dfrac{(x^2+x^3+x^4 + \hdots + x^{n+1}) - (y^2 + y^3 + y^4 + \hdots + y^{n+1})}{x-y}


    \text{The numerator has two geometric series:}

    . . S_x \:=\:x^2\cdot\dfrac{1-x^n}{1-x}\;\text{ and }\;S_y \;=\;y^2\cdot\dfrac{1-y^n}{1-y}


    \text{Therefore: }\;S_n \;=\;\dfrac{x^2\cdot\frac{1-x^n}{1-x} - y^2\cdot\frac{1-y^n}{1-y}}{x-y}

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