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Summation of a Sequence

  1. May 14, 2012 #1

    S.R

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    1. The problem statement, all variables and given/known data
    What is the sum of:
    2a2vF.png


    2. Relevant equations
    N/A


    3. The attempt at a solution
    I'm unsure how to start.

    Note: I'm in Grade 10, so I may not have the mathematical skills necessary to understand the solutions you provide.

    Any help/guidance would be appreciated.
     
  2. jcsd
  3. May 14, 2012 #2

    tiny-tim

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    Hi S.R.! :smile:

    Hint: each term is 1/√n√n+1(√n + √n+1) …

    but what is 1/(√n + √n+1) ? :wink:
     
  4. May 14, 2012 #3

    S.R

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    Assuming n+1 isn't inclusive: 1/(2√n+1). I'm not sure what to do with this information (if I'm correct, that is).
     
    Last edited: May 14, 2012
  5. May 15, 2012 #4
    Nope, n+1 is inclusive.

    The general term is

    [itex]T_n = \frac{1}{(\sqrt{n}\sqrt{n+1})(\sqrt{n} + \sqrt{n+1})}[/itex]

    But, how can you simplify this part of the above equation?

    [itex]\frac{1}{(\sqrt{n} + \sqrt{n+1})}[/itex]

    Hint:Rationalize...
     
  6. May 15, 2012 #5

    tiny-tim

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    another hint:

    nobody likes square-roots on the bottom

    nobody minds square-roots on the top :wink:
     
  7. May 15, 2012 #6

    S.R

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    Oh of course, √(n+1)-√n.
     
  8. May 15, 2012 #7
    Yep! Now, what did you do next?

    Edit : maybe its just too simple from here :biggrin:
     
    Last edited: May 15, 2012
  9. May 15, 2012 #8

    S.R

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    Simplifying the general expression, T(n)=(sqrt(n+1)-sqrt(n))^2. However, I'm stuck here. Is there an applicable formula?
     
  10. May 15, 2012 #9
    Uhh, where'd get that whole square from?? :confused:

    The simplification from the general term will yield you a difference of two terms. You can write them as

    [itex]T_1 = A_2 - A_1[/itex]
    [itex]T_2 = A_3 - A_2[/itex]

    and so on. The sum of all terms is the sum of the series. Can you notice something in the above equations that makes solving this easier?
     
  11. May 15, 2012 #10

    tiny-tim

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    nooo, Tn = (√(n+1) - √n)/√n√(n+1) = … ? :smile:
     
  12. May 15, 2012 #11

    S.R

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    Sorry I misread from my iPhone. However, the simplification from the general term is sqrt(n+1)-sqrt(n)/sqrt(n)sqrt(n+1) = sqrt(n)-sqrt(n+1).

    From your explanation I noticed, Tn=An+1-An.

    Note that the extra terms are suppose to be subscrippts.
     
    Last edited: May 15, 2012
  13. May 15, 2012 #12
    That would make [itex]2T_n = T_{n+1}[/itex] which is untrue. :wink: Its better to use different term letters for it. So, what do you get from that relation, by summing it all up?
     
  14. May 15, 2012 #13

    tiny-tim

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    no!!!!!
     
  15. May 15, 2012 #14

    S.R

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    Im unsure why not? Im not sure if my division is correct though.

    Edit: Sorry, I was replying in English class and got distracted :smile:.
     
    Last edited: May 15, 2012
  16. May 15, 2012 #15
    Its wrong :tongue2:

    [tex]Tn = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}\sqrt{n+1}}[/tex]


    Give it another go, and write out the answer :smile:
     
  17. May 15, 2012 #16

    S.R

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    Tn=1/sqrt(n)-1/sqrt(n+1)?
     
  18. May 15, 2012 #17
    Yep. Now apply the logic I suggested in post #9 and #12.
     
  19. May 15, 2012 #18

    S.R

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    I don't notice any patterns to find the sum?
     
  20. May 15, 2012 #19

    micromass

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    Can you write out the first ten terms of the sum to see if you notice anything??
     
  21. May 15, 2012 #20

    S.R

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    The sum is 9/10. Correct?
     
    Last edited: May 15, 2012
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