Hi,
one of the questions in the code4bill is as follows...
A person has to climb 22 steps.. at each point he has two choices... he can either choose to climb one step ahead or two steps ahead... you have to tell the total possible ways of climbing the steps... The problem becomes more interesting if we let the steps to be n...
When i generated solution for n using a recursive approach.. it so happens that the possible combinations for n steps is the (n+2)nd fibonacci no if we assume fibonacci starts as 0,1,1,2,3,5,8...
Can any one help me proving this pattern ??
-Karthik
