Question 923755
The way I understand it you want to know how much money should be deposited at time = 0 years,
into an account earning 8 percent annually,
so that at t=1 year, t=2 years, ..., t=8 years,
there would be at least $5000 that can be taken out.
At {{{t=0}}} , {{{balance[0] = A}}} .
At {{{t=1}}} (at the end of the first year) 8% interest ( {{{0.08A}}} is added to the balance,
bringing the balance up to {{{A+0.08A=1.08A}}} ,
and immediately, {{{5000}}} is taken out,
bringing the balance down to {{{balance[1]=1.08A-5000}}}
At {{{t=2}}} , {{{balance[1]}}} is again
increased by a factor of {{{1.08}}} ,
and reduced by the withdrawal of {{{5000}}} , so
{{{balance[2]=1.08*balance[1]-5000=1.08*(1.08A-5000)-5000=A*1.08^2-5000*1.08-5000=A*1.08^2-5000*(1.08+1)}}} .
At {{{t=3}}} , {{{balance[2]}}} is again
increased by a factor of {{{1.08}}} ,
and reduced by the withdrawal of {{{5000}}} , so
{{{balance[3]=1.08*balance[2]-5000=1.08*(A*1.08^2-5000*1.08-5000)-5000=A*1.08^3-5000*1.08^2-5000*1.08-5000=A*1.08^2-5000*(1.08^2+1.08+1)}}} .
This pattern repeats every year, and
at {{{t=10}}} ,
{{{balance[10]=A*1.08^10-5000*(1.08^9+1.08^8+1.08^7+ "..." +1.08^2+1.08+1)}}}
If at that point {{{balance[10]=0}}} ,
{{{A*1.08^10=5000*(1.08^9+1.08^8+1.08^7+ "..." +1.08^2+1.08+1)}}}
{{{A*1.08^10=5000*((1.08^10-1)/(1.08-1))}}}
{{{A*1.08^10=5000*((1.08^10-1)/0.08)}}}
{{{A=5000*(1.08^10-1)/(0.08*1.08^10)}}}
{{{A=28733.19}}}
 
NOTE:
The sum {{{SUM=1.08^9+1.08^8+1.08^7+ "..." +1.08^2+1.08+1}}}
is the sum of a geometric sequence with ratio {{{1.08}}} .
You may know a formula for that, but otherwise you can see that
{{{1.08*SUM=1.08^10+1.08^9+1.08^8+1.08^7+ "..." +1.08^2+1.08}}} , and then
{{{1.08*SUM-SUM=1.08^10-1}}}--->{{{(1.08-1)*SUM=1.08^10-1}}}--->{{{SUM=(1.08^10-1)/(1.08-1)}}}