The exercise stated that i have to solve the recurrence using the Recursion-Tree Method.
I have already finished the base part, which is $\Theta(n^{\lg3})$
But for the recursive part I'm having troubles with this sum: $cn*\sum_{i=0}^{\lg n – 1} (3/2)^i$
After resolving, I got that $T(n)=cn*n^{\lg3} – 2cn$
but this answer doesn't match $\Theta(n^{\lg 3})$ which I got using the Master Method
Best Answer
$$\sum\limits_{i=0}^{\lg n-1}(3/2)^i=\Theta((3/2)^{\lg n})\quad\text{and}\quad (3/2)^{\lg n}=2^{\lg(3/2)\lg n}=n^{\lg(3/2)}=n^{\lg3-1} $$