$T(n)=4 T(n/2) + n^2 \sqrt{2}$
I have solved this by back substitution .. and it forms equations of the form
$4k T(n/2k) + k n^2 \sqrt 2$
its giving time complexity as n2 + n2 log2n
the answer is Theta(n2.5).
i have two questions .. a) how can we get Theta(n2.5).
b) is n2 log2n Asymptotically faster than n2 ?