Why guess $\Theta(n^2)$ for the substitution method of worst-case partitioning In the book


  Introduction to Algorithms (3th edition) chapter 7


the recurrence of the running time of quicksorts partitioning is given by $$T(n) = T(n-1) + \Theta(n)$$
as the worst-case happens when the partitioning has one subproblem with $n-1$ elements and one with $0$ elements.

Now, they they guess $T(n)=\Theta(n^2)$ for the substitution method. I don't  understand this, why choose $\Theta(n^2)$? I personally would have chosen $\Theta(n)$, which is wrong.
With the help from a fellow friend I came to the following conclusion:

We know that we have to make $$n-1,n-2, \dots, 2,1 \text{ comparisons}$$ 

So when looking at the arithmetic series $$\sum_{i=1}^n k = \Theta(n^2)$$

We can conclude that $T(n) = \Theta(n^2)$. 

Please comment if I can improve my answer in any way.