I have come across this in a solution for a BMO problem where you have to find $a_{2013}$ for:
$a_n$ = $\frac{n+1}{n-1}$($a_1 + a_2 + ... + a_{n-1}$)
where $a_1$ = 1. It says that you manipulate it to get $a_n$ = $\frac {2(n+1)a_{n-1}}{n}$
However, I have no clue how he came to this. I know that you can change $a_1 + a_2 + ... + a_{n-1}$ to $\frac{(n-1)(1 + a_{n-1})}{2}$ which will give you:
$a_n = \frac{(n+1)(1 + a_{n-1})}{2}$
...but I am not sure where to go from there. Thanks, Sam.
Hint: Rearranging the recurrence relation gives:
$$\frac{n-1}{n+1}a_{n}=a_1+a_2+...a_{n-1}.$$
Substituting $n-1$ for $n$ gives:
$$\frac{n-2}{n}a_{n-1}=a_1+a_2+...a_{n-2}.$$
Now subtract the equations to get:
$$\frac{n-1}{n+1}a_{n}-\frac{n-2}{n}a_{n-1}=a_{n-1}.$$