Question 1113759
It is a geometric series, with first term {{{b=1/sqrt(3)}}} and common ratio {{{r=sqrt(3)}}} .
The sum of the first {{{n}}} terms of a geometric series with first term {{{b}}} and common ratio {{{r}}} is
{{{b((r^n-1)/(r-1))}}} .
 
In this case, the sum of the first {{{n=18}}} terms would be
{{{(1/sqrt(3))((sqrt(3)^18-1)/(sqrt(3)-1))}}}{{{"="}}}{{{(1/sqrt(3))(((3^"1 / 2")^18-1)/(sqrt(3)-1))}}}{{{"="}}}{{{(1/sqrt(3))((3^9-1)/(sqrt(3)-1))(sqrt(3)/sqrt(3))((sqrt(3)+1)/(sqrt(3)+1))
}}}{{{"="}}}{{{(19683-1)(sqrt(3)+1)sqrt(3)/(sqrt(3)sqrt(3)(sqrt(3)^2-1))}}}{{{"="}}}{{{19682(3+sqrt(3))/(3(3-1))}}}{{{"="}}}{{{19682(3+sqrt(3))/6}}}{{{"="}}}{{{highlight(9841(3+sqrt(3))/3)}}}{{{"="}}}{{{approximately15522.7}}}
 
ANOTHER WAY:
The 9 even numbered terms are
{{{matrix(1,7,
1,",",3,",",9,",...,",3^8=6561)}}} and add up to {{{1(3^9-1)/(3-1)=(19683-1)/2=9841}}} .
The 9 odd numbered terms must add up to {{{9841/sqrt(3)}}} .
Adding up both of those sums, we get
{{{9841+9841/sqrt(3)=9841(1+1/sqrt(3))=9841(sqrt(3)+1)/sqrt(3)=highlight(9841(3+sqrt(3))/3)}}}