Question 1131912
geometric series -9/2+3/2-1/2+1/6- ••• + 1/39366

{{{a[1]}}} -the first term
{{{r}}}-the "common ratio" between terms
{{{a[n]}}} -the nth term


{{{a[n]=a[1]*r^(n-1)}}}

{{{a[1]=-9/2}}} 

{{{r=(3/2)/(-9/2)}}}

{{{r=(1/1)/(-3/1)}}}

{{{r=(-1/3)}}}

{{{a[n]=-(9/2)*(-1/3)^(n-1)}}}


sum:

{{{S[n]=(a[1](1-r^n))/(1-r)}}}


{{{S[n]=(-(9/2)(1-(-1/3)^n))/(1-(-1/3))}}}...we need {{{n}}}


use  last given term  {{{1/39366}}} and plug it in


{{{a[n]=-(9/2)*(-1/3)^(n-1)}}}...last given term is {{{1/39366}}}


{{{1/39366=-(9/2)*(-1/3)^(n-1)}}}...solve for {{{n}}}


{{{(1/39366)/(-9/2)=(-1/3)^(n-1)}}}


{{{-1/177147=(-1/3)^(n-1)}}}...take log of both sides


{{{-log(1/177147)=-log((1/3)^(n-1))}}}


{{{-log(1/177147)=-(n-1)log((1/3))}}}


{{{n-1=(-log(1/177147))/(-log((1/3)))}}}


{{{n-1=log(1/177147)/log((1/3))}}}


{{{n-1=11}}}


{{{n=11+1}}}


{{{n=12}}}


=>

{{{S[12]=(-(9/2)(1-(-1/3)^12))/(1+1/3)}}}


{{{S[12]=(-(9/2)(1-1/531441))/(4/3)}}}


{{{S[12]=(-(9/2)(531440/531441))/(4/3)}}}


{{{S[12]=(-265720/59049)/(4/3)}}}


{{{S[12]=-66430/19683}}}


{{{S[12]=-3}}}{{{ 7381/19683}}}