Ultimate Question 3

this is basically a recursion problem..

in the second statement,
put i=1 j=100 k=99

you'll get f(1,100)=(1/3)f(1,99)
now put j=99,k=98(note i will always be one)
you'll get f(1,99)=(1/3)f(1,98)=(1/3²)f(1,100)
carry on....
you'll get f(1,2)=(1/3⁹⁸)f(1,100)
and f(1,2) is already (1/3)

so f(1,100)=3⁹⁷
you can easily generalise this....