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!