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- anonymous

Jason is a new college student who has saved $10000 when he was a teenager and is invested at 12 percent in a financial assent that pays interest monthly. He expects to graduate 4 years from today.
a. How much can Jason withdraw every month while he is in college if the first withdrawal occurs today ? (Answer 260.73)
b. How much can Jason withdraw every month while he is in college if he waits until the end of this month to make the first withdrawal ? ( Answer 263.34)

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- anonymous

- jamiebookeater

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- amistre64

I cant tell if the information is saying:
12% annually is divided up into monthly installments (1+(.12/12));
or if it means that the account pays out 12% each month. (1+ .12).
But if I see it right the equations go along the lines of:
a. B{n} = [B{n-1} - withdrawal] * (1 + interest)
and
b. B{n} = [B{n-1} * (1 + interest)] - withdrawal

- amistre64

i got (b) worked out :)
\begin{align}
B_n&=B_{n-1}(1.01)-w\\
&=B_{n-2}(1.01)^2-w(1.01)-w\\
&=B_{n-3}(1.01)^3-w(1.01)^2-w(1.01)-w\\
&\ or \ simply\\
B_{n}&=B_{n-3}(1.01)^3-w(1.01^2+1.01+1)\\
&=B_{n-4}(1.01)^4-w(1.01^3+1.01^2+1.01+1)\\
&=B_{n-r}(1.01)^r-w(1.01^{r-1}+1.01^{r-2}+...+1.01^2+1.01+1)\\
\end{align}
\(B_{n-r}=B_{0}\); when r=n
\begin{align}
B_{n} = B_{0}(1.01)^n - w(1.01^{n-1}+1.01^{n-2}+...+1.01^2+1.01+1)\\
\end{align}
We can clean this up by suming up those \((1.01^{n-1}...+1)\) parts like this.
\begin{align}
A&=\{\hspace{3.5em}1.01^{n-1}+1.01^{n-2}+...+1.01^2+1.01+1\}\\
(1.01)A&=\{1.01^{n}+1.01^{n-1}+1.01^{n-2+}...+1.01^2+1.01\hspace{2em}\}\\
\end{align}
\begin{align}
A&=\{\hspace{3.5em}1.01^{n-1}+1.01^{n-2}+...+1.01^2+1.01+1\}\\
-(1.01)A&=\{1.01^{n}+1.01^{n-1}+1.01^{n-2}+...+1.01^2+1.01\hspace{2em}\}\\
----&=-----------------------\\
-.01A&=1 - 1.01^n\\
----&-------\\
A &= \frac{1-1.01^n}{-.01}
\end{align}
\begin{align}
B_{n} &= B_{0}(1.01)^n -w(A)\\
B_n-B_0(1.01)^n&=-w(A)\\
\end{align}
\[w=\frac{B_n-B_0(1.01)^n}{-A}\]
\[w=\frac{B_{48}-10000(1.01)^{48}}{-A}\]
Thats one part of it :)

- amistre64

We want B{48} to equal 0 which makes this equation simply, and I use that term loosely :)
\[w = \frac{-10000(1.01)^{48}(.01)}{1-1.01^{48}}=263.338...\]

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- amistre64

And the other one I finally worked out to be:
\[w = \frac{-10000(1.01)^{48}(.01)}{1.01^{48}(.01)+1.01^{48}-1.01}=-260.73104....\]
ended up with a negative in there somehow, butI think its pretty close :)

- anonymous

Marvelous work! Thank you!

- anonymous

so from where did you get 1.01 and .01 ?

- amistre64

the 1.01 is just the interest rate of 12% compounded monthly;
1 + .12/12 = 1.01
the ".01" comes in as a result of summing up the interest over any given amount of time.
The equations start out as recursions and can be manipulated into functions with a great deal of effort :)

- amistre64

the equations are by hand so they dont follow any sort of given formula as is, but they should be close since I pretty much had to reinvent the wheel to solve them :)

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