Whenever you see $T(n)-T(n-1)=\cdots$ or equivalently $T(n+1)-T(n)=\cdots$ you should have the reflex to use a telescope.
This is because when summing, all but the first and last terms will cancel out, this is called a telescoping sum.
$\begin{array}{lcll}\require{cancel}
T(n)&-&\cancel{T(n-1)}&=a_n\\
\cancel{T(n-1)}&-&\cancel{T(n-2)}&=a_{n-1}\\
\cdots&&&\cdots\\
\cancel{T(2)}&-&T(1)&=a_2\\\hline
T(n)&-&T(1)&=\sum\limits_{i=2}^n a_i
\end{array}$
In our case this would result in:
$\displaystyle\begin{align} T(n)&=T(1)+\sum\limits_{i=2}^n (2i+1)\\&=1+2\sum\limits_{i=2}^n i+\sum\limits_{i=2}^n1\\&=1+2\left(\frac{n(n+1)}2-1\right)+(n-1)\\&=n^2+2n-2\end{align}$
You could object that you still need to prove $\sum i=\frac{n(n+1)}2$ by induction, well that's fair, though you can prove it by summing two times in reverse order $(1+2+\cdots+n)+(n+\cdots+2+1)$.
Anyway let's go back to your method and try to prove it directly by induction:
Let assume: $\,T(n)=n^2+2n-2$
$T(1)=1+2-2=1\quad\checkmark$
$T(n+1)=T(n)+2(n+1)+1=(n^2+2n-2)+2n+3=n^2+4n+1$
Now remember that we want $(n+1)^2+2(n+1)-2=n^2+2n+1+2n+2-2=n^2+4n+1$, and this is the same $\quad\checkmark$
Therefore the induction is verified.
Regarding your mistakes:
- use the simplified formula $n^2+2n-2$ instead of the fully developed one $n-1+n-n^2-1$ it is easier to handle less terms.
- in the induction you went backward, you either assume $T(n-1)$ and prove $T(n)$ or assume $T(n)$ and prove $T(n+1)$ not the opposite.
- use parenthesis, $n-1^2$ is not $(n-1)^2$ and $2n-1$ is not $2(n-1)$, this is prone to error to omit them.
Finally you can do like I did, calculate the simplified expression for $T(n+1)$ in both cases (what the end result should be, and what is it when carrying induction hypothesis) and check they are both the same. This direct calculation is easier than trying to reverse engineering it to $(n+1)^2$ and similar terms.
Best Answer
Note that $T(n) = T(n-4) + n$
Then tells us that (at least if n is a multiple of 4) that:
$$ T(n) = n + n-4 + (n-8) + ... T(0) $$
Now if we assume $T(0) = O(1)$ then we have that
$$ T(n) = n + (n-4) + (n-8) + ... O(1) $$
so we conclude that
$$ T(n) = \frac{n}{4} n + O(1) - 4 - 8 - 12 .... -n $$
Now at this point you can drop the terms on the right hand side but perhaps you are new and don't believe me so we can keep bashing this recurrence out
$$ T(n) = \frac{n}{4} n + O(1) - 4(1 + 2 + 3 ... \frac{n}{4}) $$
Now we go ahead and apply the formula that we know for sums of $1 ... n$ to find:
$$ T(n) = \frac{n^2}{4} + O(1) - 4 \frac{1}{2} \frac{n}{4} \left( \frac{n}{4} + 1 \right) $$
And this simplifies out to:
$$ T(n) = \frac{n^2}{4} + O(1) - \frac{n}{2} \left( \frac{n}{4} + 1 \right) = \frac{n^2}{4} + O(1) - \frac{n^2}{8} - \frac{n}{2}$$
Now those two quadratic terms combine but don't completely cancel out so you end up with a left over of:
$$ T(n) = \frac{n^2}{8} - \frac{n}{2} + O(1) $$
But we ignore the linear and lower terms and even forget the constant in front of the $n^2$ to just write
$$ T(n) = O(n^2) $$