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Can somene tell if I have started this right please.

  1. Apr 24, 2004 #1
    I need to find the explicit formuala for this recursive sequence.
    Code (Text):
    a[sub]1[/sub] = 3
    a[sub]2[/sub] = 7
    a[sub]n[/sub] = 7a[sub]n-1[/sub] - 10a[sub]n-2[/sub]
    a[sub]n[/sub] - 7a[sub]n-1[/sub] + 10a[sub]n-2[/sub] = 0
    t[sup]n[/sup] - 7k[sup]n-1[/sup] + 10t[sup]n-2[/sup] = 0
    t[sup]2[/sup] - 7t + 10 = 0
    (t - 2)(t - 5)
    (t = 2)(t = 5) :confused:
    if that is right I can move on.
     
    Last edited: Apr 24, 2004
  2. jcsd
  3. Apr 24, 2004 #2
    Move on. :smile:
     
  4. Apr 24, 2004 #3
    Thank you I will go ahead and stop if I have more questions I will stop. Ok?
     
  5. Apr 25, 2004 #4
    Is this the formula?
    Code (Text):
    a[sub]n[/sub] = 8/5(2)[sup]n[/sup] - 1/15(5)[sup]n[/sup]
     
  6. Apr 25, 2004 #5
    No. Probably just a mistake someplace in your algebra.

    You know, you can check your solution by using the recurrence to determine the value of a3 and comparing that to the value generated by your formula.
     
  7. Apr 25, 2004 #6
    To find B;
    Code (Text):

    2(4A + 25B = 7) = 8A + 50B = 14
    4(2A + 5B =  3) = 8A + 20B = 12
    30B = 2
    B = 1/15
     
    To find A;
    Code (Text):

    2A + 5B = 3
    2A + 1/5B = 3
    10A + 1 = 15
    10A = 16
    A = 16/10
    A = 8/5
     
    A = 8/5 B= 1/15
    ?
     
  8. Apr 25, 2004 #7
    > 2A + 5B = 3
    > 2A + 1/5B = 3 ......what is this? How did 5B become 1/5 B? You already know B. Plug it in to the first eqn & solve for A.
     
  9. Apr 25, 2004 #8
    B = 15 and A = 16?
     
  10. Apr 25, 2004 #9
    No.

    This was correct:
    Now use this equation to find A :

    2A + 5B = 3
     
    Last edited: Apr 25, 2004
  11. Apr 25, 2004 #10
    I think
    Code (Text):
    a[sub]3[/sub] = 19
     
  12. Apr 25, 2004 #11
    You think?

    You're supposed to know. :wink:
     
  13. Apr 25, 2004 #12
    Is A=1/2 if so
    Code (Text):
    the formula would be a[sub]n[/sub] = 1/2(2[sup]n[/sup]) - 1/15(5[sup]n[/sup])
     
  14. Apr 25, 2004 #13
    I don't understand what is confusing you.

    You have already solved the simultaneous equations:
    so you know that B=1/15.

    So now take one of your original equations (either one) and solve for A. Show me what you're doing, step by step.
     
  15. Apr 25, 2004 #14
    woops
    Code (Text):
    2A + 5B = 3
    2A + 5(1/15) = 3
    2A + 1 /3 = 3
    2A = 3 1/3
    A = 3 1/3 / 2
    A = 5/3
     
    Last edited: Apr 25, 2004
  16. Apr 25, 2004 #15
    [tex]B =\frac{1}{15}[/tex]
     
  17. Apr 25, 2004 #16
    2a + 5b = 3
    2a + 5(1/15) = 3
    2a + 1 /3 = 3
    2a = 3 1/3
    A = 3 1/3 / 2
    A = 5/3
     
  18. Apr 25, 2004 #17
    2a + 5b = 3
    2a + 5(1/15) = 3
    2a + 1 /3 = 3
    2a = 3 1/3 <<<<check this again
    A = 3 1/3 / 2
    A = 5/3
     
  19. Apr 25, 2004 #18
    2a + 5b = 3
    2a + 5(1/15) = 3
    2a + 1 /3 = 3
    2a = 10/3
    A = 10/3 / 2
    A = 5/3
     
  20. Apr 25, 2004 #19
    2a + 5b = 3
    2a + 5(1/15) = 3
    2a + 1 /3 = 3
    .....- 1/3 = -1/3
    2a = 8/3
    A = 4/3
     
  21. Apr 25, 2004 #20
    Thank you
    the 1/3 turns negitive? then ?
     
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