# Thread: equality problem with fractions

1. ## equality problem with fractions

hi. nice to be here.
I have to show, that

$\displaystyle \frac{n}{2n+1} + \frac{1}{(2n+1)*(2n+3)} = \frac{n+1}{2n+3}$

basicly i must confess, that i don't remember quiet a lot from school and that seems to be my problem with this. I was looking for a way to simplyfy this and my only idea was to add the two fractions. my result was:

$\displaystyle \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)} =$
$\displaystyle \frac{n+1}{2n+3}$

which -i believe- is the same as:

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{2*(2n+1)*(2n+3)}$

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{(4n+2)*(2n+3)}$

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{8n^2+12n+4n+6}$

$\displaystyle =\frac{8n^3+12n^2+2n^2+3n+2n+1}{8n^2+16n+6}$

$\displaystyle =\frac{8n^3+14n^2+5n+1}{8n^2+16n+6}$

...and basicly this is status quo, because this is, with what i have spent the last hours. i feel pleased for any answers or recommentations: this might be a trivial problem for some of you, but not for me! It's a shame, that i probably have problems with some very basic mathematical operations. please help me!

2. Originally Posted by Maya
hi. nice to be here.
I have to show, that

$\displaystyle \frac{n}{2n+1} + \frac{1}{(2n+1)*(2n+3)} = \frac{n+1}{2n+3}$

basicly i must confess, that i don't remember quiet a lot from school and that seems to be my problem with this. I was looking for a way to simplyfy this and my only idea was to add the two fractions. my result was:

$\displaystyle \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)} =$
$\displaystyle \frac{n+1}{2n+3}$

which -i believe- is the same as:

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{2*(2n+1)*(2n+3)}$

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{(4n+2)*(2n+3)}$

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{8n^2+12n+4n+6}$

$\displaystyle =\frac{8n^3+12n^2+2n^2+3n+2n+1}{8n^2+16n+6}$

$\displaystyle =\frac{8n^3+14n^2+5n+1}{8n^2+16n+6}$

...and basicly this is status quo, because this is, with what i have spent the last hours. i feel pleased for any answers or recommentations: this might be a trivial problem for some of you, but not for me! It's a shame, that i probably have problems with some very basic mathematical operations. please help me!
Without looking too closely at your work, the first step in your addition of fractions could be made a lot simpler by recognizing that the LCM of (2n+1)*(2n+3) and 2n+1 is (2n+1)*(2n+3), not (2n+1)*(2n+1)*(2n+3).

Hope that helps.

3. Originally Posted by Maya
hi. nice to be here.
I have to show, that

$\displaystyle \frac{n}{2n+1} + \frac{1}{(2n+1)*(2n+3)} = \frac{n+1}{2n+3}$

basicly i must confess, that i don't remember quiet a lot from school and that seems to be my problem with this. I was looking for a way to simplyfy this and my only idea was to add the two fractions. my result was:

$\displaystyle \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)} =$
$\displaystyle \frac{n+1}{2n+3}$

which -i believe- is the same as:

$\displaystyle =\frac{(\bold{\color{red}2}n^2+1n)*(2n+3)+2n+1}{2* (2n+1)*(2n+3)}$ <<<<<<<< typo

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{(4n+2)*(2n+3)}$

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{8n^2+12n+4n+6}$

$\displaystyle =\frac{8n^3+12n^2+2n^2+3n+2n+1}{8n^2+16n+6}$

$\displaystyle =\frac{8n^3+14n^2+5n+1}{8n^2+16n+6}$

...and basicly this is status quo, because this is, with what i have spent the last hours. i feel pleased for any answers or recommentations: this might be a trivial problem for some of you, but not for me! It's a shame, that i probably have problems with some very basic mathematical operations. please help me!
If I were you I would factor out $\displaystyle \frac1{2n+1}$ :

$\displaystyle \frac{n}{2n+1} + \frac{1}{(2n+1)*(2n+3)} = \frac1{2n+1}\left(n+\frac1{2n+3}\right)$

Now simplify the term in the bracket. Keep in mind that $\displaystyle 2n^2+3n+1=(n+1)(2n+1)$

Cancel the factor $\displaystyle (2n+1)$ to get the given result.

4. I really tried hard, but this really starts to make me sad, because i am not able to understand your answers. my problem seems to be more fundamental than i ever thought. thanks for your advise!

top 1

As I understand the hint, that "the LCM of (2n+1)*(2n+3) and 2n+1 is (2n+1)*(2n+3), not (2n+1)*(2n+1)*(2n+3)":

Let me assume, that

$\displaystyle \frac{a}{b}+\frac{c}{d} = \frac{a*d+c*b}{\color{red}{b*d}}$

so, that for

$\displaystyle \frac{n}{2n+1} + \frac{1}{(2n+1)*(2n+3)}$

i get

$\displaystyle a := n$

$\displaystyle b := (2n+1)$

$\displaystyle c := 1$

$\displaystyle d := (2n+1)*(2n+3)$

$\displaystyle \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)}$

and not

$\displaystyle \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+3)}$

top 2
i could not find a good source, which describes how to factor out something works yet. I believe this is, why i cannot see why

$\displaystyle \frac{n+1}{2n+3} = \frac1{2n+1}\left(n+\frac1{2n+3}\right)$

5. Originally Posted by Maya
hi. nice to be here.
I have to show, that

$\displaystyle \frac{n}{2n+1} + \frac{1}{(2n+1)*(2n+3)} = \frac{n+1}{2n+3}$

basicly i must confess, that i don't remember quiet a lot from school and that seems to be my problem with this. I was looking for a way to simplyfy this and my only idea was to add the two fractions. my result was:

$\displaystyle \frac{n*(2n+1)*(2n+3)+2n+1}{(2n+1)*(2n+1)*(2n+3)} =$
$\displaystyle \frac{n+1}{2n+3}$

which -i believe- is the same as:

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{2*(2n+1)*(2n+3)}$

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{(4n+2)*(2n+3)}$

$\displaystyle =\frac{(4n^2+1n)*(2n+3)+2n+1}{8n^2+12n+4n+6}$

$\displaystyle =\frac{8n^3+12n^2+2n^2+3n+2n+1}{8n^2+16n+6}$

$\displaystyle =\frac{8n^3+14n^2+5n+1}{8n^2+16n+6}$

...and basicly this is status quo, because this is, with what i have spent the last hours. i feel pleased for any answers or recommentations: this might be a trivial problem for some of you, but not for me! It's a shame, that i probably have problems with some very basic mathematical operations. please help me!
Multiply $\displaystyle \frac{n}{2n+1}$ by $\displaystyle \frac{2n+3}{2n+3}$.

This gives

$\displaystyle \frac{n(2n+3)}{(2n+1)(2n+3)} + \frac{1}{(2n+1)(2n+3)}$

These have the same denominator so you can add them

See the spoiler for why you need only multiply by $\displaystyle \frac{2n+3}{2n+3}$
Spoiler:

Consider $\displaystyle \frac{a}{b} + \frac{c}{D}$

For this special case $\displaystyle D = bd$

Hence the sum is \frac{a}{b}+\frac{c}{bd}

If we cross multipy: $\displaystyle \frac{abd+bc}{bd} = \frac{b(ad+bc)}{bd}$. Note that b will cancel and it was as though we just multiplied $\displaystyle \frac{a}{b} \cdot \frac{d}{d}$

(this is what undefined was pointing to with LCD)