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Question on finite and geometric series

  1. Jun 11, 2009 #1
    1. 1. Find the exact(no approximations)sum for the finite series

    S sub n= (2 + 2 + 2(2+...+64

    i used the parentheses to represent a radical sign

    2. Show that the sum of the first 10 terms of the geometric series

    1 + 1/3 + 1/9 + 1/27+...

    is twice the sum of the first 10 terms of the series

    1 - 1/3 + 1/9 - 1/27+...

    2. no relevant equations to solve this.

    3. I attempted to solve this by dividing n/2(a sub 1 +a sub n). n would equal 64 and a sub one would equal radical 2 so 64/2 (radical 2 +64).

    This is as far as I have been able to get and am not sure if it is the correct way to solve these two problems. I would appreciate if someone could show me step by step how to go about these since my pre-calc book does horribly at it.
  2. jcsd
  3. Jun 11, 2009 #2


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    So that is [itex]n= \sqrt{2+ 2+ 2\sqrt{2+ ...+ 64}}[/itex]?
    What does the "..." represent?

    I would think that
    [tex]\sum_{i=0}^n r^i= \frac{1- r^{n+1}}{1- r}[/tex]
    would be a very relevant equation!

  4. Jun 11, 2009 #3
    the "+...+64" is not under the radical sign. Just 2)2. and the "..." is in the question, I dont know what it stands for. Im at a complete loss with these two questions. and my math book doesnt go into enough detail with these types of equations.
  5. Jun 11, 2009 #4
    Is this [tex]S_n= \sqrt{2} + 2 + 2\sqrt{2}+...+64 = \sum_{1\le k \le 12} 2^{k/2}[/tex] ?

    if so, then it's a geometric series. Do you know the closed form of the sum of a geometric series?
    If not, heres how to get it

    [tex]S = a + ax + ax^2 + ... + ax^n = \sum_{0 \le k \le n} ax^k[/tex]

    [tex]Sx = ax + ax^2 + ax^3 + ... + ax^{n+1}[/tex]

    [tex]S - Sx = S(1-x) = a- ax^{n+1}[/tex]

    so [tex] S = a\frac{1-x^{n+1}}{1-x}[/tex]
  6. Jun 11, 2009 #5
    so the equation would not be n/2(a sub 1 + a sub n)
  7. Jun 11, 2009 #6
    for the second problem could you tell me how to find r i am unsure of how to solve for that variable
  8. Jun 11, 2009 #7
    Sorry but I don't understand what you're asking. You don't have to solve for anything, just set [tex]r=\sqrt{2}[/tex] (or [tex]x=\sqrt{2}[/tex] in my post) because [tex]a=1[/tex] in your equation.
  9. Jun 11, 2009 #8
    then i am not sure what n would be in the equation you have there.
  10. Jun 11, 2009 #9
    Well I guess that's the problem. Ellipses mean that you follow the pattern that has been set until you get to 64 (in this case).

    For example [tex]2+4+8+16+...+512[/tex] is shorthand for [tex]2+4+8+16+32+64+128+256+512[/tex].

    In your problem, the patter is that the next term in the series is generated by multiplying the previous term by [tex]\sqrt{2}[/tex]
  11. Jun 11, 2009 #10
    the final term in a geometric series can be written in the form [tex]ax^n[/tex] for some number n. The equation which you have is

    [tex]\sqrt{2} + (\sqrt{2})^2+ (\sqrt{2})^3+ ... + (\sqrt{2})^{12}[/tex]

    So n=12
  12. Jun 11, 2009 #11
    i got 216.09545 is this correct
  13. Jun 11, 2009 #12
    Yes, that's correct. Now try the other problem.
  14. Jun 11, 2009 #13
    thank you for your help
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