Running time of recursive algorithm with geometric series
$$T(n) \le 3^{k}T(\frac{n}{2^k}) + \sum_{i = 0}^k{(3/2)^i} cn$$

The recursion will reach the base case when $2^k = n$ (i.e., $k = \log n$).
And it is common to suppose the base case is $T(1) = 1$ (or $T(1) = d \textrm{ for some constant } d$).

Then (please check the details),

$$T(n) \le 3^{\log n} + \sum_{i = 0}^{\log n}{(3/2)^i} cn = n^{\log 3} + 3cn^{\log 3} - 2cn = (3c+1) n ^{\log 3} - 2cn$$