f(2n)=n+f(2n-2) and f(2)=1.
determine f(20)
This is a recurrence relation of order 2.
For a general solution, we require two initial conditions.
Do you know f(1)?
It is also possible to solve for f(20) by brute force:
given : f(2n)=n+f(2n-2) and f(2)=1
so
f(2)=1
n=2 f(4)=2+f(2)=3
n=3 f(6)=3+f(4)=6
n=4 f(8)=4+f(6)=10
n=5 f(10)=5+f(8)=15
n=6 f(12)=6+f(10)=21
n=7 f(14)=7+f(12)=28
n=8 f(16)=8+f(14)=36
n=9 f(18)=9+f(16)=45
n=10 f(20)=10+f(18)=55