Question 1041583
The sum of a geometric series is,
{{{S[n]=(a[1])((1-r^n)/(1-r))}}}
In this case, {{{a[1]=4}}} and {{{r=3}}}.
{{{4372=4(1-3^n)/(1-3)}}}
{{{4372=-2(1-3^n)}}}
{{{1-3^n=-2186}}}

{{{-3^n=-2187}}}
{{{3^n=2187}}}
{{{n=log(3,(2187))}}}
{{{n=log((2187))/log((3))}}}
{{{n=7}}}