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Simplifying the Expression

  1. Jan 16, 2016 #1
    1. The problem statement, all variables and given/known data

    Simplify:

    [tex] \frac{5\cdot 8\cdot 11 \cdots (3i+2)}{2\cdot 5 \cdot 8 \cdots (3i-1)} [/tex]

    2. Relevant equations


    3. The attempt at a solution

    I realize the numerator and denominator terms cancel besides the 2, however I'm struggling to write this in a proper form. Only just started sequences, haven't introduced infinite products or sigmas, or anything along those lines. Some insight would be appreciated.
     
  2. jcsd
  3. Jan 16, 2016 #2

    Student100

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    Just the two?
     
  4. Jan 16, 2016 #3

    vela

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    What do you mean by "proper form"?
     
  5. Jan 16, 2016 #4

    Ray Vickson

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    You say you realize that the numerator and denominator terms cancel, but I don't understand what is preventing you from just going ahead and cancelling them.
     
  6. Jan 16, 2016 #5

    I was thinking,

    [tex] \frac{3i+2}{2} [/tex]

    Because the previous term of the numerator should cancel with the 3i-1 as the pattern suggests.
     
  7. Jan 16, 2016 #6

    Student100

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    Okay, that's what you meant, not ##\frac{1}{2}##. That's it.
     
  8. Jan 16, 2016 #7
    Thanks. Been a bit sick lately; really appreciate the help on this forum.
     
  9. Jan 16, 2016 #8

    Student100

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    If you need to prove it to yourself take the first 5 terms of the sequence, and simplify. Then take i = 5 and put it into the expression you just wrote. It'll be the same. What kind of insights were you looking for?
     
  10. Jan 16, 2016 #9
    Maybe insights wasn't the proper word. I was at it a while getting no-where so I was hoping for a kick in the right direction, as was the case.
     
  11. Jan 17, 2016 #10

    HallsofIvy

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    The crucial "insight" is that 3(i- 1)+ 2= 3i- 3+ 2= 3i- 1 so that, yes, the only difference between the sums in the numerator and the denominator is that the denominator starts with "2" that the numerator does not have and that the numerator ends with 3i+ 2 that the denominator does not have. Everything else cancels.
     
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