Question 619397
There are many ways to get at the answers.
 
I calculated the common ratio (the ratio of any two consecutive terms) as
{{{r=(-1/2)/(3/2)=(-1/2)*(2/3)=-1*2/2/3=-1/3}}} {{{highlight(r=-1/3)}}}
(I chose the pair of consecutive terms so as to make my calculation so easy that that I could do in my head. It looks more complicated when I write it out, but I just asked myself what factor multiplied by 3/2 would give me -1/2, and the answer was obvious).
 
Computing that sum is not mental math, though.
We know that the sum of the first n terms in a geometric sequence is
{{{sum(a[k],k=1,k=n)=sum(a[1]*r^(k-1),k=1,k=n)=a[1]*(1-r^n)/(1-r)}}}
So for the series in the problem, substituting the values found for {{{a[1]}}, {{{r}}} and number of terms
-9/2+3/2-1/2+1/6-...+1/39366={{{sum((-9/2)*(-1/3)^(k-1),k=1,k=12)=(-9/2)* ((1-(-1/3)^12)/(1-(-1/3))) = (-3^2/2)*((1-1/3^12)/(1+1/3)))}}}
That looks ugly, let's see if we can use common denominators and simplify some powers of 3
SUM = {{{(-3^2/2)*(((3^12-1)/3^12)/(4/3))=(-3^2/2)*((3^12-1)/3^12)*(3/4)}}}={{{-3^2*(3^12-1)*3/(2*3^12*4)}}}={{{-3^3*(3^12-1)/2/4/3^12=-(3^12-1)/(8*3^9)=-(531441-1)/(8*19683)=-531440/157464}}}