Proof by the substitution method that if $T(n) = T(n - 1) + \Theta(n)$ then $T(n)=\Theta(n^2)$ 
How to prove by the substitution method that if $T(n) = T(n - 1) + \Theta(n)$ then $T(n)=\Theta(n^2)$?

I've tried the following and got stuck
$$
\begin{align}
T(n) &= T(n - 1) + \Theta(n) \\
     &= d(n - 1)^2 + cn \\
     &= d(n^2 - 2n + 1) + cn \\
     &= dn^2 - 2dn + d + cn .... \\
\end{align}$$
Any ideas?
Thanks for your time and effort!
 A: Steven has essentially given you the answer (with one minor error). Here's how one might arrive at it. I'll do one half of the derivation; the other half is the same, but you'd interchange most of the $\le$ and $\ge$ and replace every big-O with big-$\Omega$.
You know that $T(n) = T(n-1)+\Theta(n)$ so you know that $T(n) = T(n-1)+O(n)$ and so there is a $c>0$ such that 
$$T(n)\le T(n-1)+cn
$$
for all $n$ sufficiently large. We're been handed our guess, namely that $T(n)=O(n^2)$ so our guess will be that $T(n)\le kn^2$ for some $k>0$ and all $n$ sufficiently large. In the discovery phase, we'll try to find a $k$ in terms of the $c$ we have.
For our $k$ we have
$$
\begin{align}
T(n)&\le T(n-1)+cn\\
    &\le k(n-1)^2+cn
\end{align}
$$
and we require that $T(n)\le k(n-1)^2+cn\le kn^2$. We have this chain of results
$$
\begin{align}
k(n-1)^2+cn&\le kn^2\\
k(n^2-2n+1)+cn&\le kn^2\\
kn^2-2kn+k+cn&\le kn^2\\
-2kn+k+cn&\le 0\\
k(1-2n)&\le-cn\\
k(2n-1)&\ge cn\\
k&\ge \frac{cn}{2n-1}
\end{align}
$$
so since the largest value of $n/(2n-1)$ for $n\ge 1$ is $1$, any $k\ge c$ will work. Let's see that this does indeed work with $k=c$.
$$
\begin{align}
T(n)&\le T(n-1) +cn &\text{given}\\
    &\le c(n-1)^2+cn &\text{by assumption}\\
    &=cn^2-2cn+c+cn\\
    &=cn^2-cn+c\\
    &=c(n^2-n+1)\le cn^2 &\text{ta daa!}
\end{align}
$$ 
as long as $n\ge 1$, so indeed $T(n)=O(n^2)$. As I said, the other direction is immediate.
A: IMHO, a better way of phrasing the question would be: "if $g(n)\in\Theta(n)$ and $T(n)$ is defined by $T(n) = T(n-1)+g(n)$, show that $T(n)\in\Theta(n^2)$"; this stresses the role of $\Theta(n)$ as a class of functions, not an individual one.
A place to start would be to rewrite your recursive definition of $T(n)$ into a more explicit one: if you expand out a few steps of the recursion, a pattern should become clear ($T(n) = T(n-1)+g(n) = T(n-2)+g(n-1)+g(n)=\ldots$); can you figure out how to use this to write a definition of $T(n)$ that doesn't involve the recursion?
Once you have the expression from your expansion of the recursive definition of $T()$, you can start plugging in the machinery around the $\Theta()$ notation: the definition of the class $\Theta(n)$ is that $g(n)\in\Theta(n)$ if there are constants $g_1$ and $g_2$ such that for all $n$, $g_1\cdot n\leq g(n)\leq g_2\cdot n$.  The lower and upper bounds on individual values of $g()$ should let you bound the expression you found in the first step; you may want to keep in mind the formula that $\sum_{i=0}^n i = \frac{n(n+1)}{2}$.
Finally, once you have a new set of bounds for your function $T()$ you can go the other way, and try to find new constants $T_1$ and $T_2$ such that $T_1\cdot n^2\leq T(n)\leq T_2\cdot n^2$ for all $n$.  (Note that $T_1$ and $T_2$ will have to be expressed in terms of the constants $g_1$ and $g_2$ that you're guaranteed for $g()$.)  Once you have these you'll have proven your assertion that $T(n)\in\Theta(n^2)$.
To do the problem with the substitution method, a similar approach can be taken; here you'd start by picking 'constants' $T_1$ and $T_2$ (to be specified later) for the definition of $T(n)\in\Theta(n^2)$, as well as assuming the constants $g_1$ and $g_2$ specified by the problem's condition that $g(n)\in\Theta(n)$.  You can then say that
$$\begin{align}
T(n) &= T(n-1)+g(n) \\
&\leq T(n-1) + g_2\cdot n && \text{by the definition of $g(n)\in\Theta(n)$} \\
&\leq T_2\cdot (n-1)^2+g_2\cdot n&& \text{by the definition of $T(n)\in\Theta(n^2)$} \\
&=T_2 n^2+(g_2-2T_2)n+T_2&& \text{by algebra} \\
\end{align}$$
And so you need to find a way of choosing a $T_2$ such that $T_2n^2+(g_2-2T_2)n+T_2\leq T_2\cdot n^2$ for all $n$  (Hint: try subtracting $T_2n^2$ from both sides to see what you're left with).  A similar approach will handle the other side of the inequality in the $\Theta()$ definition.
