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Math Help - Sequences and Series + Natural Log

  1. #1
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    Sequences and Series + Natural Log



    I've spent so long trying to figure this out and I just CAN'T. I can see the relationship but I don't know how to actually calculate this through

    Sn = u1(r^n-1) / 4-1
    or Sn = n/2(2u1 + (n-1)d)
    I tried calculating them both as geometric and as arithmetic but I can't figure out a consistent d or r?

    Can someone help me please?

    Thank you.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by positiveion View Post


    I've spent so long trying to figure this out and I just CAN'T. I can see the relationship but I don't know how to actually calculate this through

    Sn = u1(r^n-1) / 4-1
    or Sn = n/2(2u1 + (n-1)d)
    I tried calculating them both as geometric and as arithmetic but I can't figure out a consistent d or r?

    Can someone help me please?

    Thank you.
    For this sequence to be arithmetic each term must be separated by a common difference. In other words U_3 - U_2 = U_2 - U_1

    Since \left(\frac{[\ln(2)]^3}{3!} - \frac{[\ln(2)]^2}{2!}\right) \neq \left(\frac{[\ln(2)]^2}{2!} - \frac{[\ln(2)]}{1!}\right) then the series is not arithmetic


    More it to be geometric then the ratio of each successive term must be equal:

    \frac{U_3}{U_2} = \frac{U_2}{U_1}

    \left(\frac{(\ln2) ^3}{3!} \times \frac{2!}{(\ln2) ^2}\right) \neq \left(\frac{(\ln2) ^2}{2!} \times \frac{1!}{\ln2}\right) so it is not geometric either.


    From what I can gather this is of the form \frac{(\ln2)^n}{n!} but I do not know where to go from there but hopefully someone else may be able to take it further


    edit: are you told how x and a relate to the sequence?
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  3. #3
    MHF Contributor undefined's Avatar
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    I recognize this as the Maclaurin series for f(x) = a^x with a = 2 and x =1, but I'm not sure how this might help you to solve.
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  4. #4
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    ahgh i feel so confused

    i print screened more of the worksheet

    does this additional info make any difference?

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