Let's suppose I want to resolve the following Recurrence Relation: $$ T(n) = \begin{cases} 1 & n=1 \\ T(n-1) + \Theta(n) & \text{otherwise} \end{cases} $$
I want to prove that the following $$ T(n) = \begin{cases} 1 & n=1 \\ T(n-1) + cn & \text{otherwise} \end{cases} $$ will be asymptotycally equivalent to the first Recurrence Relation.
This is how I approached the proof, but I am stuck, so I'd love to receive an help:
Proof $$ T(n) = \begin{cases} 1 & n=1 \\ T(n-1) + \Theta(n) & \text{otherwise} \end{cases} $$ means $$ T(n) = \begin{cases} 1 & n=1 \\ T(n-1) + f(n) & \text{otherwise} \end{cases} $$ where $f(n) \in \Theta(n)$.
This implies that $$\exists c_1 > 0, c_2>0, n_0>0 : c_1n \leq f(n) \leq c_2n, \ \forall n \geq n_0$$
Now let's sum $T(n-1)$: $$T(n-1) + c_1n \leq T(n-1) + f(n) \leq T(n-1) + c_2n \ \forall n \geq n_0$$
$$\implies T(n-1) + c_1n \leq T(n) \leq T(n-1) + c_2n \ \forall n \geq n_0$$
So we have $$T(n) \geq T(n-1)+c_1n\ \forall n \geq n_0$$ $$T(n) \leq T(n-1)+c_2n\ \forall n \geq n_0$$
I am close to the conclusion, but from now I cannot formally prove what I stated earlier. Thank you in advance.