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Math Help - Geometric infinite series

  1. #1
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    Geometric infinite series

    I am confused on just one review question... involving geometric infinite series. E = sigma notation.

    E(k=0 to infinity) [2(1/4)^k + 3(-1/5)^k]

    Thank you!
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  2. #2
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    okay so I was thinking seperate the two so E(k=0 to infinity) 2(1/4)^k + E(k=0 to inf) 3(-1/5)^k

    Could I pull in the 2 and 3 and make it (1/2) and (-3/5) respectively?
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  3. #3
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    Quote Originally Posted by pakman View Post
    I am confused on just one review question... involving geometric infinite series. E = sigma notation.

    E(k=0 to infinity) [2(1/4)^k + 3(-1/5)^k]

    Thank you!
    Since both series are congergent you can do each one seperately.

    First (1/4)^k is as a geometric sum: 1/(1-1/4)=4/3
    Multiplied by a factor of two infront of it: 8/3

    The second (-1/5)^5 is as a geometric sum: 1/(1+1/5)=5/6
    Multiplied by a factor of 3 is: 3/2
    When added yiedls,
    3/2+8/3
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Since both series are congergent you can do each one seperately.

    First (1/4)^k is as a geometric sum: 1/(1-1/4)=4/3
    Multiplied by a factor of two infront of it: 8/3

    The second (-1/5)^5 is as a geometric sum: 1/(1+1/5)=5/6
    Multiplied by a factor of 3 is: 3/2
    When added yiedls,
    3/2+8/3
    Thank you
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