Consider the following function:int unknown(int n) { int i, j, k=0; for (i=n/2; i<=n; i++) for (j=2; j<=n; j=j*2) k = k + n/2; return (k); }The return value of the function is

To find the return value of the function, we determine how many times the inner statement executes and how much is added to $k$ in each step:

1. Outer loop:for (i = n/2; i <= n; i++) The variable $i$ iterates from $\lfloor n/2 \rfloor$ to $n$, which is $\left(n - \frac{n}{2} + 1\right) = \Theta(n)$ iterations.

2. Inner loop:for (j = 2; j <= n; j = j*2) The variable $j$ doubles in each step starting from $2$ up to $n$ ($j = 2^1, 2^2, \dots, 2^{\lfloor \log_2 n \rfloor}$), giving $\Theta(\log n)$ iterations.

3. Loop body:k = k + n/2; In each execution of the innermost statement, the value $\frac{n}{2} = \Theta(n)$ is added to $k$.

4. Total value of $k$:

   $$\text{Total } k = (\text{outer iterations}) \times (\text{inner iterations}) \times \frac{n}{2}$$

   $$\text{Total } k = \Theta(n) \times \Theta(\log n) \times \Theta(n) = \Theta(n^2 \log n)$$

Thus, the return value of the function is $\Theta(n^2 \log n)$, which corresponds to (B).