Question 1131912
.


            The solution by  @MathLover1  seems to be almost infinitely long;
            but it can be implemented much shorter,  without making unnecessary calculations.  
            See my solution below.



<pre>
The common ratio is  q = {{{3/2}}} : {{{-9/2}}} = {{{((3/2))/((-9/2))}}} = {{{3/(-9)}}} = {{{-1/3}}}.


The formula for the sum of "n" first term of an geometry progression is


    S = {{{(a[1]*q^n-a[1])/(q-1)}}}.


We have  {{{a[1]}}} = {{{-9/2}}}  and  {{{a[1]*q^(n-1)}}} = {{{1/39366}}}  (the n-th term).


Therefore,  


    S = {{{(a[1]*q^(n-1)*q - a[1])/(q-1)}}} = {{{((1/39366)*(-1/3)-(-9/2))/((-1/3)-1)}}} = {{{(1/(3*39366) - 9/2)/((4/3))}}} = {{{((2 - 9*3*39366))/((2*3*39366))/((4/3))}}} = {{{(2-9*3*39366)/(8*39366)}}} = {{{-1062880/314928}}} = {{{-66430/19683}}} = {{{-3}}}{{{7381/19683}}}.
</pre>