Adding fractions

  • #1

Main Question or Discussion Point

(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?
 

Answers and Replies

  • #2
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##\frac{1}{3} + \frac{1}{8} = \frac{2}{4}## ? (##k = 1##)
 
  • #3
jbriggs444
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(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?
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.
 
  • #4
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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)).
Surely I can't find a common denominator with such complex variables and such?
You can always find a common denominator. Worst case: take the product of all involved denominators, that always works.
 
  • #5
Sorry, its (k / (2k + 1)) + (1 / ((2k)(2k+2))
 
  • #6
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Then it is not true in general, see post #2.
 

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