Question 328501
the formula is:
{{{S = a*((1-r^(n+1))/(1 - r))}}}
{{{a = 4}}}
{{{r = 3}}}
{{{S = 4*((1-3^(n+1))/(1 - 3))}}}
{{{S = 4*((1-3^(n+1))/(-2))}}}
I'll find out when the sum is greater than 500000
{{{S = 4*(1-3^(n+1))/(-2)}}}
{{{S = (-2)*(1-3^(n+1))}}}
{{{S/(-2) = 1 - 3^(n+1)}}}
Multiply both sides by {{{-1}}}
{{{S/2 + 1 = 3^(n+1)}}}
I can take the log of both sides
{{{log(S/2 + 1) = (n+1)*log(3)}}}
If {{{S = 50000}}}, the left side is close to {{{log(2.5) + 4}}}, so
{{{(log(2.5) + 4)/log(3) = n + 1}}}
{{{n = (log(2.5) + 4)/log(3) - 1}}}
{{{n = 4.39794/.477121 - 1}}}
{{{n = 9.21766 - 1}}}
{{{n = 8.21766}}}
Putting this back into formula,
{{{S = 4*((1-3^(n+1))/(-2))}}}
{{{S = 4*((1-3^9.21766)/(-2))}}}
{{{S = 4*(1 - 25000.08621)/(-2)}}}
{{{S = -24999.08612*(-2)}}}
{{{S = 49998.17225}}}
The next whole number for {{{n}}} is {{{9}}}
I think that would put {{{S}}} over {{{50000}}}
{{{8}}} seems to be too small