Proof

[kaebun]

prove that $\displaystyle n/(n-2)+(n+1)/(n-1)=n^2$
so $\displaystyle \frac{{n!}}{{\left( {n - 2} \right)!\left( {n-(n-2)} \right)!}} + \frac{{\left( {n + 1} \right)!}}{{\left( {n - 1} \right)!\left( {n+1-(n-1)} \right)!}}$
then $\displaystyle \frac{{n!}}{{\left( {n - 2} \right)!\left( {2!} \right)}} + \frac{{\left( {n + 1} \right)!}}{{\left( {n - 1} \right)!\left( {2!} \right)}}$
$\displaystyle \frac{{n!}}{{\left( {n - 2} \right)!\left( {2} \right)}} + \frac{{\left( {(n + 1)n} \right)!}}{{\left( {(n - 1)!n} \right)\left( {2} \right)}}$
$\displaystyle \frac{{n!}}{{\left( {n - 2} \right)!\left( {2} \right)}} + \frac{{\left( {(n + 1)n} \right)!}}{{\left( {n!} \right)\left( {2} \right)}}$

I'm not sure where to go from here
thanks for any advice

 

[pka]

Sorry to say, but I cannot read your post.
What are you to prove? What is “the punch line” of the proof?

When using TeX try to use the fraction notation.
EX: \(\displaystyle \L
\frac{{n!}}{{\left( {n - 2} \right)!\left( {2!} \right)}} + \frac{{\left( {n + 1} \right)!}}{{\left( {n - 1} \right)!\left( {2!} \right)}}\)
is made up as \frac{{n!}}{{\left( {n - 2} \right)!\left( {2!} \right)}} + \frac{{\left( {n + 1} \right)!}}{{\left( {n - 1} \right)!\left( {2!} \right)}}

 

[kaebun]

is that better?

 

[pka]

Yes the formating is better!
But you did no tell us what to prove.
You just listed some statements.

 

[kaebun]

that n/(n-2)+(n+1)/(n-1)=n^2

 

[pka]

IT IS NOT TRUE! \(\displaystyle \L
\frac{n}{{n - 2}} + \frac{{n + 1}}{{n - 1}},\quad n = 3\quad \frac{n}{{n - 2}} + \frac{{n + 1}}{{n - 1}} = 5\)

 

[kaebun]

oh i'm sorry i forgot this part n is less than or equal to 2