we can also go by this method i got in a stanford's coursera course
Now given is that a(alpha) is the probability of getting the filled blocks , thus (1-a) is the probability of getting the empty blocks.
now let E(N) be the expectation of the average number of probe required when inserting an element .
Thus E(N) comes out to be 1 + (a*E(N)).
here 1 is added for the operation of going and probing while "a" is multiplied so that when the probe comes out. to be having an element already , it can be done all over again and thus E(N) is multiplied
solving E(N)=1 + (a*E(N))
E(N)=1/(1-a). the required answer.