Answer will be $\Theta(n)$
$j = n/2+n/4+n/8+\ldots +1$
$\quad = n \left[1/2^1 + 1/2^2 + 1/2^3 +\ldots + 1/2^{\lg n}\right] $
(Sum of first $n$ terms of GP is $\left[a . \frac{1-r^n}{1-r}\right],$ where $a$ is the first term, $r$ is the common ratio $< 1,$ and $n$ is the number of terms)
$\quad = n \left[1/2 \frac{1 - (1/2)^{\lg n}}{1-1/2} \right]$
$\quad = n \left[\frac{n-1}{n}\right]$
$\quad = n-1 = \Theta(n)$