High School Adding Fractions: Simplifying ((k/2k+1)+(1/(2k)(2k+2))=((k+1)/(2k+2))

[PhysicsBoyMan]

(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?

 

[fresh_42]

$\frac{1}{3} + \frac{1}{8} = \frac{2}{4}$ ? ($k = 1$)

 

[jbriggs444]

(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.

 

[mfb]

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.

 

[PhysicsBoyMan]

Sorry, its (k / (2k + 1)) + (1 / ((2k)(2k+2))

 

[mfb]

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