>>9881826The way to check that it is correct is to see that (3/2)^n - 1 is 0 when n=0 and that it satisfies induction equation as E[X_n], ie. (3/2)^{n+1} - 1 = 3/2((3/2)^n - 1) + 1/2
Now, that is not how you actually come up with it. To get the formula, note that by applying the induction equation several times, you get:
E[X_n] = 3/2 E[X_{n-1}] + 1/2 = 3/2 (3/2 E[X_{n-2}] + 1/2) + 1/2 = ...
Now with some imagination, it is not very hard to guess that after n steps you get:
E[X_n] = (3/2)^n E[X_0] + (3/2)^{n-1} * (1/2) + (3/2)^{n-2}* 1/2 + ... + 1/2 = ((3/2)^{n-1} + (3/2)^{n-2} + ... + 1) * 1/2 = (3/2)^n - 1
