How to go from a recurrence relation to a final complexity
From what you wrote it seems that for all $k$ you have $T(k)=k-1+T(k-1)$ and $T(1)=0$. Therefore you can get it directly:

$$T(n)=(n-1)+(n-2)+\dots+1+0 = \sum_{k=1}^{n}{(k-1)}=\frac{n(n-1)}{2}$$

So $T(n)$ is in $Θ(n^2)$.