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Yahoo!

  • 2 years ago

Find the n th term and sum of n terms of the series 5 + 8 + 17 + 44 ..........

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  1. dpflan
    • 2 years ago
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    Ah, this is a nice series. \[[5,8,17,44]\]

  2. experimentX
    • 2 years ago
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    find the difference .. note the difference in the difference is a multiple of 3

  3. Yahoo!
    • 2 years ago
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    Yes the first order difference is in Gp

  4. dpflan
    • 2 years ago
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    and on...so do as @experimentX suggests, first examine any sort of relationship that could exist, you see that each termer is greater than the next

  5. dpflan
    • 2 years ago
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    I suggest that you show us the values you have for the difference between the numbers in the series

  6. dpflan
    • 2 years ago
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    Once, you have those differences, see if there is a pattern or relationship among them

  7. zepp
    • 2 years ago
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    Looks like 3 to some power :D

  8. experimentX
    • 2 years ago
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    tn+1 = tn + 3^(n-1)

  9. dpflan
    • 2 years ago
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    term1 = 5 term2 = 8 term2 - term1 = 3 so term2 = term1 + 3 term3 = 17 term3 - term2 = 9 so term3 = term2 + 9 term4 = 44 term5 - term4 = 27 so term4 = term3 + 27 term5 = ...

  10. experimentX
    • 2 years ago
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    don't know closed form at the moment!!

  11. dpflan
    • 2 years ago
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    So @Yahoo! How are you doing with this?

  12. Yahoo!
    • 2 years ago
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    Ar^n + b1n^k-1 + b2n^k-2............bk

  13. dpflan
    • 2 years ago
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    Hmm, can you explain the terms please?

  14. Yahoo!
    • 2 years ago
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    it is the general formula lol1

  15. sami-21
    • 2 years ago
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    ok look at the sequence first term is (3^1+7)/2=5 second is (3^2+7)/2=8 third is (3^3+8)/2=17 can you now tell me general formula ?

  16. experimentX
    • 2 years ago
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    i guess we should use summation rule |dw:1343147827410:dw|

  17. mahmit2012
    • 2 years ago
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    |dw:1343147906405:dw|

  18. mahmit2012
    • 2 years ago
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    |dw:1343147929358:dw|

  19. experimentX
    • 2 years ago
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    not sure that works!

  20. experimentX
    • 2 years ago
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    |dw:1343147968035:dw| something like that!! just try to manipulate and fit in it ...

  21. mahmit2012
    • 2 years ago
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    |dw:1343147962155:dw|

  22. experimentX
    • 2 years ago
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    |dw:1343148038487:dw|

  23. mahmit2012
    • 2 years ago
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    |dw:1343148023475:dw|

  24. amistre64
    • 2 years ago
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    i like to get the first term to 1 alot of times 5 + 8 + 17 + 44 -4 -4 -4 -4 ----------------- 1 + 4 + 13 + 40 just helps me with the inbetween addings or if we up it by +1 5 + 8 + 17 + 44 +1 +1 +1 +1 ------------------- 6 + 9 + 18 + 45

  25. amistre64
    • 2 years ago
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    factor out a 3 3( 2 + 3 + 6 + 15 + ...) all ways to se it differently

  26. experimentX
    • 2 years ago
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    |dw:1343148174027:dw|

  27. experimentX
    • 2 years ago
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    i hope it works for other terms too .. let's see for sum

  28. Yahoo!
    • 2 years ago
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    @experimentX i didnt understand!!

  29. experimentX
    • 2 years ago
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    |dw:1343148311125:dw|

  30. experimentX
    • 2 years ago
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    that's just some manipulation .. typing in latex really takes lot of time

  31. experimentX
    • 2 years ago
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    \[ t_n = t_{n-1} + 3^{n-1}\] this is the general rule \[ t_{n} = t_{n-2} + 3^{n -2} + 3^{n-1} \] continue this way and you will get \[ t_{n} = t_1 + 3 + 3^3 + 3^4+ .. + 3^{n -2} + 3^{n-1} = 5 + 3{ 3^{n-1} - 1 \over 3-1}\] which is requited formula

  32. experimentX
    • 2 years ago
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    for summation \[ \sum_{n=1}^k {5 + {3 \over 2} (3^{n - 1 } - 1)} =\sum_{n=1}^k{7/2} + 3^n = k {7 \over 2}+ 3{3^k -1 \over 3-1} = k {7 \over 2}+ {3 \over 2}(3^k -1)\] @Yahoo! do you understand??

  33. mahmit2012
    • 2 years ago
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    |dw:1343149457407:dw|

  34. mahmit2012
    • 2 years ago
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    @experimentX I guess you missed 1/2 .

  35. experimentX
    • 2 years ago
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    thanks @mahmit2012 for correction ... you are sharp!!ii

  36. Yahoo!
    • 2 years ago
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    @mahmit2012 @experimentX i learnt this by using this Ar^n + b1n^k-1 + b2n^k-2............bk

  37. Yahoo!
    • 2 years ago
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    formula!1

  38. mahmit2012
    • 2 years ago
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    You're welcome , and good tutoring you have had. @experimentX

  39. Yahoo!
    • 2 years ago
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    Ar^n + b1n^k-1 + b2n^k-2............bk

  40. Yahoo!
    • 2 years ago
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    can u ssolve this by using Ar^n + b1n^k-1 + b2n^k-2............bk

  41. experimentX
    • 2 years ago
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    you are a good tutor!! most of me is autodidact!!

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