How to show that $T(n) = T(n-1) + \Theta(n)$ is in $\Omega(n^2)$ In the class we have been shown the way to prove that 
$T(n) = T(n-1) + \Theta(n)$ is in $O(n^2)$

$$
\begin{align}
T(n)&amp;\le T(n-1) +cn &amp;\\
    &amp;\le c(n-1)^2+cn &amp;\\
    &amp;=cn^2-2cn+c+cn\\
    &amp;=cn^2-cn+c\\
    &amp;=c(n^2-n+1)\le cn^2 &amp;\
\end{align}
$$ 

Then we were said that it is easy to see that $T(n) = T(n-1) + \Theta(n)$ is in $\Omega(n^2)$. But, I do not understand how to "see" this really.
As I understand we need to show that

$$
T(n-1) + \Theta(n)\ge cn^2
$$
Right? I have no idea how to do that. Need help. Thank you!
Let $c$ be the lower bound so that your $\Theta (n)$ is greater than or equal to $cn$. Then you have

$$T(n) \geq T(1) + c\sum_{j=1}^n j = T(1) + cn(n+1)/2 = \Omega(n^2)$$

The only tricky thing there is knowing the formula for $\sum_{j=1}^n j$. But that's a classical result, you can look up sum of consecutive integers for example.
The following definition is useful for your purpose  $f(n)=\Theta(g(n))$ means that 

$$ c_1\, g(n) \leq  f(n) \leq c_2 \,g(n) $$  

for some positive $k_1,k_2$.