Question 802064
The sides of each triangle will be {{{1/2}}} as long as the sides of the prior triangle, so the first perimeter is {{{3}}}, the second is {{{3/2}}}, the next {{{3/4}}}, and so on.
The perimeters form a geometric sequence.
The sum converges to {{{highlight(6)}}}.
That is easy to see because as you add each perimeter you are reducing the difference between your sum and {{{6}}} by a factor of {{{2}}}.
 
If you need to show your work, you could dust all those formulas about geometric sequences, and use the version taught in your class.
 
Maybe you would use the fact that the sum of the first {{{n}}} terms of a geometric sequence is
{{{S[n]=a[1](1-r^n)/(1-r)}}}
where {{{a[1]}}} is the first term and {{{r}}} is the common ratio.
In this case {{{a[1]=3}}} and {{{r=1/2}}}, so
{{{S[n]=3(1-(1/2)^n)/(1-(1/2))=3(1-(1/2)^n)/(1/2)=2*3(1-(1/2)^n)=6(1-(1/2)^n)}}}
As {{{n}}} increases without bounds, {{{(1/2)^n}}} approaches zero, and {{{s[n]}}} approaches {{{highlight(6)}}}.
 
Maybe you were taught that if {{{r<1}}} an infinite geometric series converges to {{{a[1]/(1-r)}}}
In that case, plugging {{{a[1]=3}}} and {{{r=1/2}}} into that formula you get that the sum of the infinite series is {{{3/((1/2))=3*2=highlight(6)}}}