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anonymous

  • 5 years ago

i am baffled about how to do the following problem Sum to n terms the following series 2/(1.3.5) + 4/(3.5.7) + 6/(5.7.9) ...... I FOund this in a textbook in a paragraph about summing series by the method of differences.

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  1. anonymous
    • 5 years ago
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    need you amistre

  2. amistre64
    • 5 years ago
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    the top is even sums right?

  3. amistre64
    • 5 years ago
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    the bottom is cycling thru odds

  4. amistre64
    • 5 years ago
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    ....same problem btm? ;)

  5. anonymous
    • 5 years ago
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    yes

  6. amistre64
    • 5 years ago
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    k, give me a minute :)

  7. anonymous
    • 5 years ago
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    yes top and bottom are AP's

  8. anonymous
    • 5 years ago
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    k thanks!

  9. amistre64
    • 5 years ago
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    n=1; k=2 n=2; k=4 n=3; k=6 top is 2n

  10. amistre64
    • 5 years ago
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    n=1; a=1.b=3.c=5 n=2; a=3.b=5.c=7 n=3; a=5.b=7.c=9 if we solve fo "a" then ther rest are just add ons

  11. anonymous
    • 5 years ago
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    yes the general fortmula of the nth term = 2n- top part that is

  12. amistre64
    • 5 years ago
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    the bottom is 2n-1 for a

  13. amistre64
    • 5 years ago
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    2n --------------- 2n-1.2n+1.2n+3

  14. anonymous
    • 5 years ago
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    nth term for btm is(2n-1)(2n+1)(2n+3)

  15. amistre64
    • 5 years ago
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    at n=4 8 ------ 7.9.11

  16. amistre64
    • 5 years ago
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    times? if thats what the "." means then sure :)

  17. anonymous
    • 5 years ago
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    right - but the problem is we need to find the sum of n terms - this is where i got stuck

  18. anonymous
    • 5 years ago
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    yes '.' means multiply

  19. amistre64
    • 5 years ago
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    well, we have the equation; all we need to do is figure out the summation for it I spose

  20. amistre64
    • 5 years ago
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    2n 2n ------------------ = -------------- (2n-1)(2n+1)(2n+3) (4n^2-1)(2n+3)

  21. anonymous
    • 5 years ago
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    yeah right - I' must find out what the 'differences ' method is about. I think this stuff is really advanced.

  22. amistre64
    • 5 years ago
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    2n ------------------ 8n^3 +12n^2 -2n -6

  23. amistre64
    • 5 years ago
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    n ---------------- is what I get for a basis 4n^3 +6n^2 -n -3

  24. anonymous
    • 5 years ago
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    yes - thats right

  25. anonymous
    • 5 years ago
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    - but where do we go from here?

  26. amistre64
    • 5 years ago
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    we could split this by partial fraction decomposition if you think that would help to find seperate summations for each part...

  27. anonymous
    • 5 years ago
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    good idea

  28. amistre64
    • 5 years ago
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    which the first equation would be more helpful :)

  29. amistre64
    • 5 years ago
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    2n A B C ----------------- = ----- + ------ + ------ (2n-1)(2n+1)(2n+3) (2n-1) (2n+1) (2n+3)

  30. anonymous
    • 5 years ago
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    yep

  31. amistre64
    • 5 years ago
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    2n = A(2n+1)(2n-1) + B(2n-1)(2n+3) + C(2n-1)(2n+1)

  32. amistre64
    • 5 years ago
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    rewrite, thats miss typed lol 2n = A(2n+1)(2n+3) + B(2n-1)(2n+3) + C(2n-1)(2n+1)

  33. anonymous
    • 5 years ago
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    right 2nd time

  34. amistre64
    • 5 years ago
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    when n = 1/2; we get B0 + C0 + A(2)(4) = 1; A = 1/8

  35. amistre64
    • 5 years ago
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    when n = -1/2 we get A(0) + C(0) + B(-2)(2) = -1; B=1/4

  36. amistre64
    • 5 years ago
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    when n = -3/2; we get B0 + A(0) +C(-4)(-2) = -3; C = -3/8 right?

  37. amistre64
    • 5 years ago
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    do you see how I got those numbers? and if so did I do it right :)

  38. anonymous
    • 5 years ago
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    yes i understand your working - let me check them

  39. anonymous
    • 5 years ago
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    yes -that ok i think

  40. amistre64
    • 5 years ago
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    inthe meantime; we will also need to factor out the "2" from the denominators to get the "n" all by itself

  41. amistre64
    • 5 years ago
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    1 1 3 ------ - ------- - -------- 8(2n-1) 4(2n+1) 8(2n+3)

  42. amistre64
    • 5 years ago
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    1 1 3 --------- - --------- - -------- 16(n-1/2) 8(n+1/2) 16(n+3/2)

  43. amistre64
    • 5 years ago
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    Now to sum to the "nth" we can sum each seperately...\[[\frac{1}{16}\sum_{}\frac{1}{n-\frac{1}2}{}] -[\frac{1}{8}\sum_{}\frac{1}{(n+\frac{1}{2})}] -[\frac{1}{16}\sum_{}\frac{1}{n+\frac{3}{2}}]\]

  44. amistre64
    • 5 years ago
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    there is a process to get to the iterations; f(xi) delta(xi)

  45. amistre64
    • 5 years ago
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    (b-a)/n is the form im familiar with; where a is the leftmost in the interval and b is the rightmost. and n - the number of partitions....

  46. amistre64
    • 5 years ago
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    but thats starting to get into the cobwebs in my memory :)

  47. amistre64
    • 5 years ago
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    if n = the interval 1 to 10 and we partition it out every 1 unitl that would be 9; so I am thinking our delta(xi) = n-1

  48. amistre64
    • 5 years ago
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    any of this sound familiar or reasonable :)

  49. amistre64
    • 5 years ago
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    lets test my memory here: suppose we want to sum the numbers 1-9; that should equal 45 the equation would simply be "n" n(n-1) = n^2 -n our summation would amount to: n(n+1)(2n+1) n(n+1) ------------ - ------- if my head is right :) 6 2

  50. amistre64
    • 5 years ago
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    9(10)(19) 9(10) -------- - ----- = 1704 -45 = 1659; lol... i guess im rusty :) 6 2

  51. anonymous
    • 5 years ago
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    I got the same 3 sum expressions as you did amis but came to a dead end. i believe there must be a different approach to solve this one.

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