We first get the nth term of the sequence

3,6,12,...

That's a geometric series with a1 = 3, r = 2

The nth term of that sequence is a∙rn-1 or 3∙2n-1

So the log series is



We write each number in parentheses as a product of 3 and some
other number:

 

Now we use the principle that the log of a product equals the sum
of the logs of the factors:



Since there are n terms, there and n log(3)'s added so the above 
sequence is:



We write each number, 1,2,4,...2n-1 in the parentheses are powers of 2.



Use the principle of logs that says that the log of an exponential is the
exponent times the log of the base:



We factor out log(2)



The series in parentheses is an arithmetic series with a1 = 0,
common difference 1 and number of terms n.  We use the sum formula for
the series in parentheses:







Substituting for the series in parentheses:







Edwin