Solve the Recurrence Relation $T(n) = 2T(n/2) +T(n/4) + n^2$. I am trying to find the best way to solve $T(n)=2T(\frac{n}{2})+T(\frac{n}{4})+n^2$. I tried to use the recurrence tree method, and after adding up all the levels of the tree i get the summation below:
$T(n)=n^2+\frac{9n^2}{16}+\frac{18n^2}{64}+\frac{36n^2}{256}+...$
When I try to find a closed form solution I get something like $288n^2$. Can I say that the recurrence is $\Theta(n^2)$?
 A: To make things simple, we denote that $n = 2^k$ ($k$ is not necessarily $\in \mathbb{N}$) and the recurrent sequence becomes
$$T(2^k) = 2T(2^{k-1}) + T(2^{k-2}) + 2^{2k} \tag{1}$$
Let's denote $f(k) = T(2^k)$, then
$$\begin{align}
(1)&\Longleftrightarrow f(k) = 2f(k-1) +f(k-2) + 4^{k} \\
&\Longleftrightarrow \left(f(k) - \frac{16}{7}4^k\right) = 2\left(f(k-1) - \frac{16}{7}4^{k-1}\right) +\left(f(k-2) - \frac{16}{7}4^{k-2}\right)  \tag{2}\\
\end{align}$$
Denote $g(k) = f(k) - \frac{16}{7}4^k$, then
$$(2)\Longleftrightarrow g(k) = 2g(k-1) +g(k-2) \tag{3}$$
$(3)$ is a well-know recurrent sequence, so it's easy to prove that
$$g(k) = Ax_1^k+Bx_2^k \tag{4}$$
with $(x_1, x_2) = ( 1 + \sqrt{2},  1 - \sqrt{2})$ are two roots of the quadratic equation $x^2=2x+1$ and $A,B$ can be determined from the initial values of the sequence $g(k)$ (for example, $g(0)$ and $g(1)$).
Finally, we have
$$T(2^k) =f(k) =g(k)+\frac{16}{7}4^k = Ax_1^k+Bx_2^k + \frac{16}{7}4^k$$
or with $k = \ln(n) / \ln(2)$
$$\begin{align}
T(n) &= Ax_1^\frac{\ln(n)}{\ln(2)}+Bx_2^\frac{\ln(n)}{\ln(2)} + \frac{16}{7}4^\frac{\ln(n)}{\ln(2)} \\
&=A n^\frac{\ln|x_1|}{\ln 2} + B n^\frac{\ln|x_2|}{\ln 2} + \frac{16}{7}n^2\tag{5}\\
\end{align} $$
We observe that $2 > \frac{\ln|x_1|}{\ln 2} > \frac{\ln|x_2|}{\ln 2}$, so the third term of $(5)$ is dominant for $n \longrightarrow + \infty$. Then
$$T(n) \sim \frac{16}{7}n^2 + \mathcal{o}(n^2)$$
