Question 998671
The sum of the first {{{2}}} terms, {{{a[1]=1}}} and  {{{a[2]=8}}} , is
{{{a[1]+a[2]=1+8=9=3^2}}} .
The sum of the first {{{3}}} terms, going all the way to {{{a[3]=18}}} , is
{{{a[1]+a[2]+a[3]=9+a[3]=9+18=27=9^3}}} .
The sum of the first {{{4}}} terms, going all the way to {{{a[4]}}}= the fourth term (to be found), is
{{{a[1]+a[2]+a[3]+a[4]=27+a[4]}}} must be {{{3^4=81}}} .
So {{{27+a[4]=81}}}-->{{{a[4]=81-27}}}-->{{{a[4]=highlight(54)}}} .


EXTRA:
If {{{a[1]+a[2]+"..."+a[n-1]+a[n]=3^n}}} ,
then {{{a[1]+a[2]+"..."+a[n-1]=3^(n-1)}}} , and
{{{a[n]=(a[1]+a[2]+"..."+a[n-1]+a[n])-(a[1]+a[2]+"..."+a[n-1])}}}
{{{a[n]=3^n-3^(n-1)}}}
{{{a[n]=3*3^(n-1)-3^(n-1)}}}
{{{a[n]=(3-1)*3^(n-1)}}}
{{{a[n]=2*3^(n-1)}}}
All of that for {{{n>1}}} , of course, as the problem specified.
For {{{n=1}}}, the sum of the first {{{n=1}}} terms is {{{a[1]=1}}} ,
which is not equal to {{{3^1=3}}} .
That formula {{{a[n]=2*3^(n-1)}}} is how I mentally calculated {{{a[4]=2*3^3=2*27=54}}} ,
because {{{81-54}}} is the type of calculation that I have hated to do since the first grade.