can someone help me

[chad brad]

i have tries values of k (0,1,2,3,4,5) and none of the answers matched the outcomes of f(2^k) matched the answers given.

 

[Dr.Peterson]

View attachment 22646

i have tries values of k (0,1,2,3,4,5) and none of the answers matched the outcomes of f(2^k) matched the answers given.

You must have made a mistake.

What did you get for f(1), f(2), f(4), f(8), f(16), and f(32)? Did you see any pattern that might lead to a general formula?

What did you get for each of the options for n = 1, 2, 4, 8, 16, 32?

 

[JeffM]

It is impossible to see where you have gone astray, if you have gone astray, because you show no work.

Actually, all of the proposed answers work if k = 0.

[MATH]k \in \mathbb Z^+ \implies n = 2^k > 1 \text { and } \dfrac{n}{2} \in \mathbb Z \implies [/MATH]
[MATH]\left \lfloor \dfrac{n}{2} \right \rfloor = \dfrac{n}{2} \implies f(n) = 2 * f\left(\dfrac{n}{2}\right) + 1 = n + 1.[/MATH] ...................[edited]

So you are correct. The answers do not work.

In fact for any even integer x, the result will be x + 1.

 

[HallsofIvy]

The crucial point is that $\displaystyle 2^k$, as you repeatedly divide by 2, is always an integer (so you don't have to worry about the "integer" part) and always a power of two.

For example if k= 4 (and $\displaystyle x= 2^4= 16$) then f(x)= 2f(8)+ 1. But f(8)= 2f(4)+ 1, f(4)= 2f(2)+ 1, f(2)= 2f(1)+ 1, and f(1)= 0. So f(2)= 2(0)+ 1= 1, f(4)= 2(1)+ 1= 3, f(8)= 2(3)+ 1= 7, and f(16)= 2(7)+ 1= 15.
Hey, that's just $\displaystyle x-1$! I could try to prove that works for all x BUT I see that n-1 is one of the answers and none of the other answers are 15 for n= 16. Ah, the joys of "multiple guess" questions!