Recurrence - finding asymptotic bounds for $T(n) = T(n-2) + n^2$ I've been working on a problem set for a bit now and I seem to have gotten the master method down for recurrence examples. However, I find myself having difficulties with other methods (recurrence trees, substitution). here is the question I am stuck on:
$$T(n) = T(n-2) + n^2$$
Is there a pattern as follows?
$$n^2 + T(n-2) + T(n-4) +...$$ 
where it goes until there is no more n left. so around n/2 times
and would that mean that 
$$n^2 + (n-2)^2 + (n-i) ^2$$ so the asymptotic bound would be $\theta(n^2)$?
I am honestly taking a shot in the dark here, so I was hoping someone could help guide me in how to approach these questions.
Thank you,
Tyler

$$T(n) = T(n-2) + n^2 = T(n-4) + (n-2)^2 + n^2 = T(n-2k) + \sum\limits_{i = 0}^{k - 1}(n - 2i)^2$$
This goes down while $n - 2k \ge 0$. Assuming even $n$ (for asymptotic complexity, it does not really matter, and you can do similar calculations for odd $n$ also, with the same asymptotic results), we have $k = \frac{n}{2}$ at the end.
$$T(n) = T(0) + \sum\limits_{i = 0}^{\frac{n}{2} - 1}(n - 2i)^2 = \sum\limits_{i = 0}^{\frac{n}{2} - 1}(n^2 - 4ni + 4i^2) + C$$
$$T(n) = n^2\cdot\left(\frac{n}{2}-1\right) - 4n\cdot\frac{1}{2}\cdot\frac{n}{2}\cdot\left(\frac{n}{2} - 1\right) + 4\cdot\frac{1}{6}\cdot\left(\frac{n}{2} - 1\right)\cdot\frac{n}{2}\cdot(n-1) + C$$
$$\therefore \ T(n) = \Theta(n^3)$$
