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

Note that if $n=2k$ is even, then
$$
T(n)+T(n-1) = n^2+(n-1)^2+ \cdots+4^2+3^2 + T(2)+T(1) =\frac{n(n+1)(2n+1)}{6} + C.
$$
Here $C=T(2)+T(1) -2^2-1^2$ and we used the formula $\sum_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6}$.
We also note that $T(n) \sim T(n-1)$, so we may conclude that 
$$
T(n) \sim n^3/12.
$$
