Unless i am mistaken somewhere (which is always possible),
T(n) = 2*T(N/2) + 2  : not + 1.. there are 2 comparisons.

the equation should simplify as follows:
let n = 2^k;

T(n) = 2^(k-1)+2^(k-1) + ... + 8 + 4 + 2
     = 2^(k-1)+21+2+4+ ... + 2^(k-2)
     = 2^(k-1)+2(2^(k-1) - 1/2-1)
     = 2^(k-1) + 2^k - 2
     = 2^(k-1)/2 + 2^k - 2
     = 3*2^k/2 - 2
     =  3n/2  -  2

not that it REALLY matters inthe grander scheme of things ...
but i think that is right ...