Not able to find any pattern for $4T(n/2)+n^2 n^{1/2}$
Here is what happens when you expand the recurrence:
$$
\begin{align*}
T(n) &amp;= n^{5/2} + 4(n/2)^{5/2} + 4^2(n/2^2)^{5/2} + 4^3(n/2^3)^{5/2} + \cdots \\ &amp;=
n^{5/2} \left(1 + \frac{4}{2^{5/2}} + \left(\frac{4}{2^{5/2}}\right)^2 + \left(\frac{4}{2^{5/2}}\right)^3 + \cdots \right).
\end{align*}
$$
Since $4/2^{5/2} = 1/\sqrt{2} &lt; 1$, the large bracketed expression converges, and so the answer is $T(n) = \Theta(n^{5/2})$.