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B Adding fractions

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  1. Jul 20, 2016 #1
    (k / 2k + 1) + (1 / (2k)(2k+2)) = ((k+1) / (2k+2))

    I would like to simplify the left side to prove that these two statements are equal. I'm not sure how to do this. Surely I can't find a common denominator with such complex variables and such? What is a good approach?
     
  2. jcsd
  3. Jul 20, 2016 #2

    fresh_42

    Staff: Mentor

    ##\frac{1}{3} + \frac{1}{8} = \frac{2}{4}## ? (##k = 1##)
     
  4. Jul 20, 2016 #3

    jbriggs444

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    A good first step would be to explain what (k / 2k + 1) is supposed to mean.

    Is it supposed to denote ##\frac{k}{2k + 1}## or is it supposed to denote ##\frac{k}{2k} + 1## ? The latter is what it does denote according to the PEMDAS rules.
     
  5. Jul 20, 2016 #4

    mfb

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    2016 Award

    Staff: Mentor

    It is unclear how to interpret your fractions.
    (k / 2k + 1) = k / (2k + 1)? Probably what you meant.
    (k / 2k + 1) = (k / (2k)) + 1? More logical given the usual operator order (multiplication/division before addition)
    (k / 2k + 1) = (k / 2)k + 1 = (k2/2) + 1? That's how a computer would interpret it.

    Same thing for (1 / (2k)(2k+2)).
    You can always find a common denominator. Worst case: take the product of all involved denominators, that always works.
     
  6. Jul 20, 2016 #5
    Sorry, its (k / (2k + 1)) + (1 / ((2k)(2k+2))
     
  7. Jul 20, 2016 #6

    mfb

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    2016 Award

    Staff: Mentor

    Then it is not true in general, see post #2.
     
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