\problemdata      
        {annuities-3} % latex-friendly label for the prob.
        {Annuities} % The topic(s) of the problem content.
        {?} % Source (if known)
        {S02 PS6; F01 PS6} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem}
Suppose you deposit \$100 into your MIT Credit Union account
today, \$99 in one month from now, \$98 in two months from now, and so on.
Given that the interest rate is constantly 0.3\% per month, how long will
it take to save \$5,000?

\solution{
First note that you will certainly manage to have saved \$5,000
\emph{some} day, since, even without your earnings from the interest,
you will have $100+99+\cdots+1=100\times 101/2 = 5,050$ dollars after
99~months. 

But fewer months will be needed: After the first deposit you will have
\$100. After the second deposit, you will have \$$(100\times 1.003 +
99)$. After your third deposit, your saved money will be \$$\bigl(
(100\times 1.003 + 99)\times 1.003 + 98) =$ \$$\bigl( 100\times
(1.003)^2 + 99\times 1.003 + 98\bigr)$, and so on. So, after the $n$th 
deposit,  
$$
S_n = \sum_{i=0}^{n-1} (100 - i) (1.003)^{n-i-1}
$$
dollars will be in your account. Substituting $j=n-i-1$, we can
rewrite this as  
$$
\sum_{j=0}^{n-1} \bigl( 100 - (n-j-1)\bigr) (1.003)^{j};
$$
and then as 
$$
(101-n) 
\Bigl( \sum_{j=0}^{n-1} (1.003)^{j}   \Bigr)+
\Bigl( \sum_{j=0}^{n-1} j (1.003)^{j} \Bigr).
$$
Using the closed forms from Course Notes~6, we can finally write $S_n$
as 
\[
(101-n)
\Bigl( \frac{1-1.003^n}{1-1.003} \Bigr)+
\Bigl( \frac{1.003-n1.003^n+(n-1)1.003^{n+1}}{(1-1.003)^2} \Bigr).
\]
Solving 
$$
S_n \geq 5,000
$$ 
for $n$, we get $n\geq 67$. That is, you'll need more than 5.5~years
to save~\$5,000.
}
\end{problem}

