Question 1013535
<pre>
{{{log((3))+log((6))+log((12))+""*""*""*""}}}

We first get the nth term of the sequence

3,6,12,...

That's a geometric series with a<sub>1</sub> = 3, r = 2

The nth term of that sequence is a&#8729;r<sup>n-1</sup> or 3&#8729;2<sup>n-1</sup>

So the log series is

{{{log((3))+log((6))+log((12))+""*""*""*""+log((3*2^(n-1)))}}}

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

{{{log((3*1))+log((3*2))+log((3*4))+""*""*""*""+log((3*2^(n-1)))}}} 

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

{{{(log((3))^""+log((1)))+(log((3))^""+log((2)))+(log((3))^""+log((4)))+""*""*""*""+(log((3))^""+log((2^(n-1))))}}}

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

{{{n*log((3))+(log((1))^""+log((2))+log((4))+""*""*""*""+log((2^(n-1))))}}}

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

{{{n*log((3))+(log((2^0))^""+log((2^1))+log((2^2))+""*""*""*""+log((2^(n-1))))}}}

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

{{{n*log((3))+(0*log((2))^""+1*log((2))+2*log((2))+""*""*""*""+(n-1)*log((2)))}}}

We factor out log(2)

{{{n*log((3))+log((2))*( 0^""+1+2+""*""*""*""+(n-1) )}}}

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

{{{S[n]}}}{{{""=""}}}{{{expr(n/2)(a[1]^""+a[n])}}}

{{{S[n]}}}{{{""=""}}}{{{expr(n/2)(0+(n-1)^"")}}}

{{{S[n]}}}{{{""=""}}}{{{expr(n/2)(n-1)}}}

Substituting for the series in parentheses:

{{{n*log((3))+log((2))*( 0^""+1+2+""*""*""*""+(n-1) )}}}

{{{n*log((3))+log((2))*( expr(n/2)(n-1) )}}}

{{{n*log((3))+n(n-1)log((2))/2}}}

Edwin</pre>