Find Big O using Iteration
You're almost there. You correctly have
$$\begin{align}
T(n) &amp;= T(n-1)+2n\\
     &amp;= [T(n-2)+2(n-1)] + 2n = T(n-2)+2(n-1)+2n\\
     &amp;= T(n-3)+2(n-2)+2(n-1)+2n\\
     &amp;= T(n-4)+2(n-3)+2(n-2)+2(n-1)+2n
\end{align}$$
and in general,
$$
T(n)=T(n-j)+2(n-j+1)+2(n-j+2)+2(n-j+3)+\dotsm+2n
$$
Now all you need is the base value for the recurrence. Let's suppose that it was $T(0)=0$ (for the big-O estimate it really wouldn't matter if we had had, say $T(1)=3$ as our base).

To get to the base case, $T(0)$, we'd then need $n-j=0$. In that case, we'd have
$$
T(n)=T(0)+2(1)+2(2)+2(3)+\dotsm+2n= \sum_{k=1}^n2k=2\cdot\frac{n(n+1)}{2} = O(n^2)
$$