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Arithmetic Sequence

  1. Jan 2, 2012 #1
    1. The problem statement, all variables and given/known data
    V0=4

    [tex]V_{n+1}=\sqrt{V_{n}^{2}+2n+3}[/tex]

    2. Relevant equations
    Show that Un is an arithmetic sequence.


    3. The attempt at a solution
    I counted Vn and i found that it equals:

    [tex]V_{n}=\sqrt{(Vn+2)^{2}+2}[/tex]

    what is there to do after this?
     
  2. jcsd
  3. Jan 2, 2012 #2

    SammyS

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    What is Un? Is that a typo, or is Un = (Vn)2 ?
     
  4. Jan 2, 2012 #3
    No there is no Un at all.
     
  5. Jan 2, 2012 #4

    SammyS

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    That looks like a Un to me.

    BTW: How do you count Vn ?
     
  6. Jan 4, 2012 #5
    Sorry it's show that Vn is arithmetic
     
  7. Jan 4, 2012 #6

    Curious3141

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    Well, it clearly can't be, because [itex]V_0 = 4[/itex], [itex]V_1 = \sqrt{19}[/itex] and [itex]V_2 = \sqrt{24}[/itex] and [itex]V_2 - V_1 \neq V_1 - V_0[/itex] establishing that there is no common difference.
     
  8. Jan 4, 2012 #7
    sorry V0=1

    V0=1 V1=√6 V2=√11 V3=√16


    I found that Un=1+√(1+5n)

    And i know that arithmetic series are written as Un=Up+nr

    so: Up=1 and r=5 therefore you get: Un=1+√(1+5n)

    is that all I have to do?
     
  9. Jan 4, 2012 #8

    Curious3141

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    What is [itex]U_n[/itex]? You've only defined what [itex]V_n[/itex] is so far. :grumpy:
     
  10. Jan 4, 2012 #9

    Curious3141

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    By asking what [itex]U_n[/itex] is, I don't mean just quote a formula which you've derived. Please define exactly what [itex]U_n[/itex] is supposed to represent.

    It might be better if you reproduced the exact question in its original form, word for word.
     
  11. Jan 4, 2012 #10
    well the general form of an arithmetic series is :

    Un=Up+nr and in my case Un is Vn and Up is V0 and r is 5 so i get:

    [tex]V_{n}=1+\sqrt{1+5n}[/tex]
     
  12. Jan 4, 2012 #11

    Curious3141

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    OK, it's all clearer now. Part of the confusion lay in the fact that you had miscalculated the terms for [itex]V_n[/itex] repeatedly.

    Forget about trying to fit things into a particular form for now. Let's start by looking at some values of [itex]V_n[/itex], correctly computed.

    Please recalculate [itex]V_1, V_2, V_3[/itex] and [itex]V_4[/itex] very carefully, and you'll see a much simpler pattern emerging. We'll take it from there.
     
  13. Jan 4, 2012 #12

    Curious3141

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    After you do that, you will need to find a closed form expression for [itex]V_n[/itex] (the one you previously derived is clearly wrong), then formally prove it with mathematical induction. Once that's done, it'll become immediately apparent that [itex]V_n[/itex] is the general term of an arithmetic progression (in fact, one of the simplest and most well-known arithmetic progressions).
     
  14. Jan 4, 2012 #13
    alright i'll count them right now.
     
  15. Jan 4, 2012 #14
    i keep getting V0=1 V1=√6 V2=√11 V3=√16

    idk what is wrong?
     
  16. Jan 4, 2012 #15

    Curious3141

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    V0 = 1 (given)

    V1 = sqrt(1 + 2*0 + 3) = sqrt (4) = ?

    Once you get V1 wrong, the rest will be wrong too, so restart from here.

    You were probably doing sqrt(1+2*1 + 3), but remember the index for the initial term is zero.
     
  17. Jan 4, 2012 #16
    Wow I didn't pay attention to that sorry:

    V0=1 V1=2 V2=3 V3=4 and so one

    so the general difference is 1
     
  18. Jan 4, 2012 #17
    Yes I didn't pay attention to that.
     
  19. Jan 4, 2012 #18
    Vn in terms of n is Vn=1+n
     
  20. Jan 4, 2012 #19
    To check I did Vn+1-Vn=2+n-1-2=1

    so the common difference is 1.

    do i also have to so Vn+2-Vn+1?
     
  21. Jan 4, 2012 #20

    Curious3141

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    No need. You've now got an expression for [itex]V_n[/itex]. You need to prove it.

    Two ways.

    First is a direct proof, which might proceed like so:

    [tex]V_{n}^2 = V_{n-1}^2 + 2(n-1) + 3[/tex]

    [tex]V_{n-1}^2 = V_{n-2}^2 + 2(n-2) + 3[/tex]

    ...

    [tex]V_1^2 = V_0^2 + 2(0) + 3[/tex]

    then successively substituting the equation below into the one above until one gets:

    [tex]V_{n}^2 = V_0^2 + 2(\frac{1}{2})(n-1)(n) + 3n[/tex]

    which can be simplified to:

    [tex]V_{n}^2 = {(n+1)}^2[/tex]

    [tex]V_{n} = n+1[/tex]

    Fairly simple. But I would recommend the second method, mathematical induction. Try and do this as an exercise, and post your results here.
     
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