i'm normally decent with these type of problems but i can't really figure this one out.
show that if n is a power of 2, say n = 2^k, then
$\displaystyle \sum_{i=0}^{k} log(n/2^i) = \Theta (lg^2 n)$
if n is a power of two then $\displaystyle \sum_{i=0}^{k} log(n/2^i) = \sum_{i=0}^{k} log(2^{k-i}) =\sum_{i=0}^{k} (k-i)log(2)$ $\displaystyle = log(2) \sum_{i=0}^{k} (k-i) \in \Theta(k^2) = \Theta([log_2(n)]^2) $ for some $\displaystyle k = log_2(n)$