smc

Let's establish some ground rules... $a=$ The first term in the geometric sequence. $r=$ The ratio relating the terms of the geometric sequence. $n=$ The nth value of the geometric sequence, starting at 1 and increasing as consecutive integer values. Using these terms, the sum can be written as: $sum=a/(1-r)=3$ Let $x=$ The positive integer that is in the reciprocal of the geometric ratio. This gives: $3=a/(1-(1/x))$ $3=ax/(x-1)$ Now through careful inspection we notice that when x = 3 the equation becomes $3=a(3/2)$, where $a=2$. Therefore $r=1/x=1/3$. Now we define the sum as $2\ast (1/3{)}^{n-1}$. Now we simply add the $n=1$ and $n=2$ terms. $sum=2(1/3{)}^{(1)-1}+2(1/3{)}^{(2)-1}=2+2/3=6/3+2/3=8/3$ This gives $\boxed{\frac{8}{3}}$.