Let's take a simple example.
For convenience and simplicity let's take the code as,
for(i = 1; i < = n; i++)
Now trace the program on paper manually.
You will find that For i=1 ,
Hello will be Printed 4 times.
For i=2 ,
Hello will be Printed 2 times.
For i=3 ,
Hello will be Printed 1 time.
After Loop Gets Terminated
Finally parent Process(Original One) will Print Hello.
That is 8 times Hello will be printed, in which only last one belongs to the Parent Process.
So, total Child Processes = 8-1(Parent)= 7 = 23 - 1.
total Child Processes = 2n-1(Parent) = 2n - 1.