Mathematics of Finance

Question

If you invest a dollar at “6% interest compounded monthly,” it amounts to $1.005^n$ dollars after $n$ months. If you invest 10 dollars at the beginning of each month for 10 years (120 months), how much will you have at the end of the 10 years? I know the answer is $1646.99 but do not know how to get this answer using the compound interest formula. Please explain.

Answer 1

Your first deposit compounds for 120 months. The second for 119 months. Your last deposit for 1 month.

The future value of 120 deposits.

$10(1.005)^{120} + 10(1.005)^{119} + \cdots + 10(1.005)$

or

$\sum_{n=1}^{120} 10(1.005)^n$

This is the sum of a geometric progression.

$\sum_{n=1}^{m} y^n = \frac {y(y^m-1)}{y-1}$

to find this formula multiply by $\frac {1-y}{1-y}$

$\frac {1}{1-y}(1-y)(y+y^2 + y^3 + \cdots + y^m) = \frac {1}{1-y} (y - y^2 + y^2 -y^3 + y^3-y^4 + \cdots - y^{m+1})$

The expression "telescopes" leaving:

$\frac {1}{1-y} (y - y^{m+1})$ which equals the formula above.

plugging the numbers for this problem.

$10\frac {1.005(1.005^{120} - 1)}{0.005}$