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If you invest $\$100$ per month at an annual return of $15\%$ and kept reinvesting the profits, after:

$$19 \text{ years, you would have }$119,000.00$$

$$ 50 \text{ years, you would have } $9,600,000.00$$

$$ \text{ 60 years, you would have } $39,200,000.00$$

$$ \text{ 70 years, you would have } $158,000,000.00$$

It is possible to convert this into a formula so I can enter in the year to give me the total, i.e. $10$ years would = ?, $100$ years would =?

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If you have a total profit of $P_n$ after $n$ years, how much do you have after $n+1$ years in terms of the previous $P_n$? –  anon Aug 2 '11 at 13:50

1 Answer 1

up vote 2 down vote accepted

Let $P(n)$ be the principal at the end of $n$ months. I will presume the interest is paid monthly at the end of the month and you invest first at month 1. Then $P(1)=100\ \ $ and each month you pay interest on the existing balance and add $100$, so $P(n)=1.0125P(n-1)+100.\ \ \ $ This can be solved by the usual techniques for recurrence relations.

Another approach is to note that each $100$ invested grows independently of the others. So $P(n)=\sum_{i=1}^n100(1.0125)^{n-i},$ where $i$ refers to the month of the deposit and each term in the sum is what that deposit is worth at month $n$. This is a geometric series and can be summed as such, giving $P(n)=100\frac{1.0125^{n}-1}{.0125}$

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