Calculate Geometric Progression with logarithm

Question

I'm trying to find the following summation but cant seem to find the answer. Would really appreciate any help from here!

$$\sum_{i=0}^{k}(4/3)^{i}\log(n/3^i)$$

Answer 1

$$(4/3)^i\log(n/3^i)=(4/3)^i(\log n-\log(3^i))=(4/3)^i(\log n-i\log 3)$$

So, your sum is,

$$\log n\left[\sum_{i=0}^k (4/3)^i\right]-\log 3\left[\sum_{i=0}^k i(4/3)^i\right]$$

The first summand has a finite geometric series and the second summand has a finite arithmetico-geometric series. Do you know how to compute them?