Question 1097575
{{{39366=2*3^9}}} , so {{{1/39366=1/(2*3^9)=3^(-9)/2}}} .
The series is
{{{-3^2/2+3^1/2-3^0/2+.3^(-1)/2+"..."+3^(-9)/2=sum((-9/2)*(-1/3)^k,k=0,11)}}} .
It is a geometric series with 12 terms,
first term {{{b=-9/2}}} and common ratio {{{r=-1/3}}} .
The sum of the first {{{n}}} terms of a geometric series with
first term {{{b}}} and common ratio {{{r}}} is
{{{S[n]=b(r^n-1)/(r-1)}}} .
In this case, blindly applying that formula, we would calculate
{{{S[12]=(-9/2)((-1/3)^12-1)/(-1/3-1)=(-9/2)((1/3)^12-1)/(-4/3)=(9/2)(3/4)(1/3^12-1)=(9*3/(2*4))((1-3^12)/3^12)=(3^3/2^3)*((1-3^12))/3^12=(1/2^3)*((1-3^6)(1+3^6)/3^9)=(1/8)*(1-729)(1+729)/19683=(1/8)(-728*730/19683)=-91*730/19683=-66430/19683}}}
 
A SMARTER WAY:
If we pair those 12 terms, we have 6 terms
{{{(-9/2+3/2)+(-1/2+1/6)+"..."=-3+-1/3+"..."}}}
forming a geometric series with first term {{{-3}}} and ratio {{{1/9}}}
The sum then can be calculated as
{{{(-3)((1-(1/9)^6)/(1-1/9))=(-3)(((9^6-1)/9^6)/(8/9)))=-3((9^6-1)/9^6)(9/8)=-3((9^3-1)(9^3+1)/9^6)(9/8)}}}={{{-3*728*730/(8*9^5)}}}={{{-3*91*730/9^5=-91*730/3^9=-66430/19683}}}