Question

spose you got $15000 in an account that pays 8% interest compounded monthly; how much can you withdrawal to make the account last no more than 3 years?

Answer 1

withdrawal each month to make the account go zero after 3 years is a better wording :)

Answer 2

A = P(!-r)^t

Answer 3

1 st Month = P (1 + R) - W
2 nd Month = ((P (1 + R) - W) (1 + R)) - W
\[= P(1+R)^2-W(1+R)-W\]
3rd Month =\[P(1+R)^3-W(1+R)^2-W(1+R)-W\]
\[Nth Month =P(1+R)^N-W\left(1+(1+R)+(1+R)^2+\text{...}.(1+R)^{N-1}\right)\]

Using Geometric Sum Formula
\[ \left(1+(1+R)+(1+R)^2+\text{...}.(1+R)^{N-1}\right)=\frac{\left(1-(1+R)^N\right)}{1-(1+R)}=-\frac{1-(1+R)^N}{R}\]

Putting it back together:
\[ P(1+R)^N-W\left(-\frac{1-(1+R)^N}{R}\right)=P(1+R)^N+W\frac{\left(1-(1+R)^N\right) }{R}\]
P- Initial amount
R- monthly interest rate
N= months
W= Withdrawal
Plugging and Setting the equation to zero and Finding W

\[15000\left(1+\frac{.08}{12}\right)^{36}+W\frac{\left(1-\left(1+\frac{.08}{12}\right)^{36}\right) }{\frac{.08}{12}}==0,\]
W=470.045

Answer 4

Excel agrees with my result