Show, by induction, that $T(n) = \frac{n(n+1)}{2}$ given the defined piecewise function. I have no idea how to solve this. My math proving skills are pretty rusty.  The problem gives the following definition to start:
$$T(n) = \begin{cases}
 1 \text{ if } n = 1 \\
 T(n-1) + n \text{ otherwise}
\end{cases}$$
Show, by induction, that $T(n) = \frac{n(n+1)}{2}$.
My Attempt
So they're giving a recurrence relation as a place to start.  So, I imagine I need to find a closed form.  Expanding the piecewise function we have,
\begin{align*}
T(1) &= 1 \\
T(2) &= T(2-1) + 1 \\
     &= T(1) + 1 \\
     &= 1 + 1 \\
     &= 2 \\
T(3) &= T(3-1) + 1 \\
     &= T(2) + 1 \\
     &= 2 + 1 \\
     &= 3 \\
     &\vdots \\
T(n) &= \frac{n(n+1)}{2} \text{ since it appears to be the sum of the first n ints.}
\end{align*}
Proof.
Base case.  Let $n = 1$.  Then, $T(1) = \frac{1(1+1)}{2} = \frac{2}{2} = 1$, as defined.
Induction.  Suppose, $T(n) \implies T(n+1)$.  Then by definition,
$$
\frac{n(n+1)}{2} \implies \frac{(n+1)((n+1)+1)}{2} = \frac{(n+1)(n+2)}{2} = \frac{n^2+3n+2}{2} 
$$
Suppose $\frac{n(n+1)}{2}$ for some $n > 1$.
This is as far as I get.  I know I need to use the inductive hypothesis at some point here, I just don't know/remember how/where to do that at.  Any help is greatly appreciated.
 A: As I've indicated in the comments, some of your proof doesn't make any sense linquistically, and thus logically. Probably this is because induction was explained to you using a functional notation for the statements (almost certainly "$P(n)$" because for some reason "$P$" is the favored letter for this). And here you see a function $T(n)$ and are confusing it with the statement notation. But $T(n)$ is not a statement. It is just a number. The corresponding statement that is represented by $P(n)$ is "$T(n) = \frac{n(n+1)}2$".

So they're giving a recurrence relation as a place to start.  So, I imagine I need to find a closed form.

No. They gave you a closed form in the problem. You don't need to find it. You just need to prove that it actually is equal to the function.

$$\begin{align*}
T(2) &= T(2-1) + 1 \\
&\ \ \vdots\\
T(3) &= T(3-1) + 1\end{align*}$$

Not quite: $T(2) = T(2-1)+2, T(3) = T(3-1)+3$.
To prove a set of statements $P[n]$ for each $n \ge 1$ by (simple) induction, you prove the base case: $P[1]$, and the induction case: "$P[n] \implies P[n+1]$", or in words: for any $n$, if the statement $P[n]$ is true, then the statement $P[n+1]$ is also true.
For this problem the base case $P[1]$ is "$T(1) = \frac{1(1+1)}2$", for which your current proof is fine.
The induction case is "$P[n] \implies P[n+1]$", which is $$\left(T(n) = \frac{n(n+1)}2\right) \implies \left(T(n+1) = \frac{(n+1)(n+2)}2\right)$$
And you do not "suppose" this. This is the statement you are trying to prove. This is an implication, and the normal approach for proving an implication is to suppose that the hypothesis of the implication is true, and from that statement you then derive the conclusion.
That is, you suppose $$T(n) = \frac{n(n+1)}2$$
And then from this statement, you derive
$$T(n+1) = \frac{(n+1)(n+2)}2$$
It is that derivation that proves the implication
$$\left(T(n) = \frac{n(n+1)}2\right) \implies \left(T(n+1) = \frac{(n+1)(n+2)}2\right)$$
