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Math Help - Series using summation notation

  1. #1
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    Series using summation notation

    Series is 4 + 2 + 1/2 + 1/4 + 1/8

    My work:

    6 sigma n=1 4(1/2)^n-1

    Could it be; 6 sigma m=1 2^n-1 as well?


    I'm a little confused, the teachers solutions are:

    5 sigma n=0 2^2-n

    I've no idea how he got this...
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  2. #2
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    Quote Originally Posted by thekrown View Post
    Series is 4 + 2 + 1/2 + 1/4 + 1/8

    My work:

    6 sigma n=1 4(1/2)^n-1

    Could it be; 6 sigma m=1 2^n-1 as well?


    I'm a little confused, the teachers solutions are:

    5 sigma n=0 2^2-n

    I've no idea how he got this...
    Did you plan to include 1 in the series or is it deliberately omitted ....?
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  3. #3
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    What do you mean?

    As far as I know when you start at n=1 then you have to put a"-1" after the n on the right side after sigma and when you start with n=0 you don't need to.

    The exponent part of my teachers answer is beyond me. Any clue what he did there, or if my answer is good? He noted that other answers are possible but only listed the one I wrote above.
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  4. #4
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    Quote Originally Posted by thekrown View Post
    What do you mean?

    As far as I know when you start at n=1 then you have to put a"-1" after the n on the right side after sigma and when you start with n=0 you don't need to.

    The exponent part of my teachers answer is beyond me. Any clue what he did there, or if my answer is good? He noted that other answers are possible but only listed the one I wrote above.
    your series isn't complete, probably...
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  5. #5
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    Smile

    if your series is,4 + 2 +1+1/2 + 1/4 + 1/8 then you might have,
    \sum_{k=0}^{5}4(\frac{1}{2})^k
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  6. #6
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    Perfect, I got that before I tried to morph it into what the teacher had (and failed).

    Originally I had what you had except above sigma was 6, below was 1 and on the right I had n-1.

    Is either version just as good?
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  7. #7
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    Smile

    Quote Originally Posted by thekrown View Post
    Perfect, I got that before I tried to morph it into what the teacher had (and failed).

    Originally I had what you had except above sigma was 6, below was 1 and on the right I had n-1.

    Is either version just as good?
    it's exactly the same.
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  8. #8
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    Okay thank you.
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  9. #9
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    Smile

    Remember !
    \sum_{k=1}^{n}x_k=\sum_{k=1+i}^{n+i}x_{k-i}
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  10. #10
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    Quote Originally Posted by thekrown View Post
    Series is 4 + 2 + 1/2 + 1/4 + 1/8 [snip]
    Quote Originally Posted by mr fantastic View Post
    Did you plan to include 1 in the series or is it deliberately omitted ....?
    Quote Originally Posted by thekrown View Post
    What do you mean?

    As far as I know when you start at n=1 then you have to put a"-1" after the n on the right side after sigma and when you start with n=0 you don't need to.

    The exponent part of my teachers answer is beyond me. Any clue what he did there, or if my answer is good? He noted that other answers are possible but only listed the one I wrote above.
    Quote Originally Posted by Raoh View Post
    if your series is,4 + 2 +1+1/2 + 1/4 + 1/8 then you might have,
    \sum_{k=0}^{5}4(\frac{1}{2})^k
    @OP: This was my point in post #2. If you had been paying closer attention you might have realised that ....
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