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
 A: $$T(n) = T(n-2) + n^2 = T(n-4) + (n-2)^2 + n^2 = T(n-2k) + \sum\limits_{i = 0}^{k - 1}(n - 2i)^2$$
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.
$$T(n) = T(0) + \sum\limits_{i = 0}^{\frac{n}{2} - 1}(n - 2i)^2 = \sum\limits_{i = 0}^{\frac{n}{2} - 1}(n^2 - 4ni + 4i^2) + C$$
$$T(n) = n^2\cdot\left(\frac{n}{2}-1\right) - 4n\cdot\frac{1}{2}\cdot\frac{n}{2}\cdot\left(\frac{n}{2} - 1\right) + 4\cdot\frac{1}{6}\cdot\left(\frac{n}{2} - 1\right)\cdot\frac{n}{2}\cdot(n-1) + C$$
$$\therefore \ T(n) = \Theta(n^3)$$
A: 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 
$$
T(n) \sim n^3/12.
$$
A: The fact that
$T(n) = T(n-2) + n^2$
has the values
for even and odd $n$
being completely independent
suggests
(at least to me)
that we
consider separately
even and odd $n$.
For even $n$,
let
$R(n) = T(2n)$,
so,
since
$T(2n)
=T(2n-2)+(2n)^2
$,
$R(n)
=R(n-1)+(2n)^2
$.
Therefore,
$R(n)
=R(0)+\sum_{k=1}^n (2k)^2
=R(0)+4\frac{n(n+1)(2n+1)}{6}
\approx 4n^3/3
$
so,
if $n$ is even,
$T(n)
= R(n/2)
\approx n^3/6
$.
Similarly,
for odd $n$,
let
$S(n) = T(2n+1)$
so,
since
$T(2n+1)
=T(2n-1)+(2n+1)^2
$,
$S(n)
=S(n-1)+(2n+1)^2
$.
Therefore,
$\begin{array}\\
S(n)
&=S(0)+\sum_{k=1}^n (2k+1)^2\\
&=S(0)+\sum_{k=1}^n (4k^2+4k+1)\\
&=S(0)+\sum_{k=1}^n 4k^2+\sum_{k=1}^n 4k+\sum_{k=1}^n 1\\
&=S(0)+4\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}+n\\
&\approx 4n^3/3\\
\end{array}
$
so,
if $n$ is odd,
$T(n)
= S((n-1)/2)
\approx n^3/6
$
again.
A: let $n=log m$
$$T(\log m)=T(\log m-\log4)+(\log m)^2$$
$$T(\log m)=T(\log m/4)+(\log m)^2$$
$$s(m)=s(m/4)+m^2$$
here $a=1,b=4$ and $f(n)=m^2$
$$m^{\log a}=1$$
$$f(m)>m^{\log a}$$ (base 4)
case (iii) of masters theorem follows
hence $s(m)=\Theta(m^2)$
after comouting $c$ through $af(m/b)=cf(n)$
$$c=1/16<1$$
hence case (iii) totally followed
and $s(m)=\Theta(m^2)$
and $T(m)=\Theta(2^2n)$
m not confirm about the answer but the procedure will be the same
