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Homework Help: Discrete math set theory sum problem!

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if k>1 then,

    5/(k-1)-3/k-2/(k+2) = (9k+6)/(k-1)k(k+2)

    Hence simplify Ʃ of k=2 to n {(3k+2)/(k-1)k(k+2)}

    2. Relevant equations

    3. The attempt at a solution

    Ok so the first part is ok I just multiplied the denominators with the numerators and expanded and cancelled.

    However the second part (hence simplify).... is where im really stuck I have no clue what to do... please help! Thanks
  2. jcsd
  3. Mar 15, 2012 #2


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    hi charmedbeauty! :smile:

    write the LHS sums separately …

    (and use some spaces!! :rolleyes:)

    ∑ 5/(k-1) - ∑ 3/k - ∑ 2/(k+2) ​

    and then change the variable in two of the sums :wink:
  4. Mar 15, 2012 #3
    Alternatively, you can write out the first few (I would recommend at least 8) sums of the series (the series that Tim wrote out) and see if you can spot of pattern of terms canceling. Most telescoping series problems work out nicely with canceling; the answer for this one isn't pretty looking, but it is solvable!

    And, just out of curiosity, you meant to type

    Hence simplify [itex]\displaystyle\sum_{k=2}^{n} \frac{9k+6}{(n-1) \cdot n \cdot (n+2)}[/itex], correct?
  5. Mar 15, 2012 #4
    Thanks, but no... all the terms in the denominator are in terms of k and not n!
  6. Mar 15, 2012 #5
    Oops! I meant for the terms in the denominator to be k! I was referring to your inconsistency between the 3k+2 in the numerator and the 9k+6 in the numerator. Which is it? Have you worked out a solution yet?
  7. Mar 15, 2012 #6
    its (3k + 2) that needs to be simplified. the (9k +3) refers to the first part of the question! so somehow these are related im guessing (obviously its a factor of 3 although i dont know if this has any significance.)

    But to clarify The first part of the question asks to simply the expression in terms of a single denominator you do this and you get (9k +3) as the numerator.
    The next part has a DIFFERENT numerator, namely, (3k+2).

  8. Mar 15, 2012 #7


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    hi charmedbeauty! :smile:
    no, the first numerator was (9k+6), which is exactly 3 times (3k+2) :wink:
  9. Mar 16, 2012 #8
    Sorry, yes that is correct.

    Anyway I have worked it out now ......


    thanks for the input.
  10. Mar 16, 2012 #9
    Your constant term is a little bit off, but otherwise the other terms of your answer are correct. You probably made a little error when adding up your constant terms, that's all.

    Also, I know what you meant by your answer, but in the future use parentheses around the denominator of fractions so it avoids confusion.

    [itex]1/3(55/6-5/n-2/n+1-2/n+2)= \displaystyle\frac{1}{3} \cdot \left(\frac{55}{6} - \frac{5}{n} - \frac{2}{n} + 1 - \frac{2}{n} + 2\right) \neq \frac{1}{3} \cdot \left(\frac{55}{6} - \frac{5}{n} - \frac{2}{n+1} - \frac{2}{n+2}\right)[/itex]

    Like I mentioned earlier, the constant term is off, retry the calculation and see you messed up!
  11. Mar 16, 2012 #10
    Last edited: Mar 16, 2012
  12. Mar 16, 2012 #11
    Is the last expression not the answer?
  13. Mar 16, 2012 #12
    The constant term is a little bit off. The terms involving the n variables are correct. Just check to see if you added up the constant terms correctly.
  14. Mar 16, 2012 #13

    hmm just did it with a calculator still getting 55/6
  15. Mar 16, 2012 #14
    The first terms that don't cancel are: [itex]\frac{1}{3} \cdot \left(\frac{5}{1} - \frac{3}{2} + \frac{5}{2} - \frac{3}{3} + \frac{5}{3}\right) = \frac{20}{9}[/itex]
  16. Mar 16, 2012 #15
    Ohhh I thought the terms with three in the numerator cancelled.
  17. Mar 17, 2012 #16
    Doesnt the 3/3 cancel out with the 5/5.?
  18. Mar 17, 2012 #17
    I didn't get any [itex]\frac{5}{5}[/itex] term when I carried the partial sums out.
  19. Mar 17, 2012 #18
    No sorry my mistake I checked it again not realising that the 5/4 went with the -3/4 and -2/4
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