Problem 125 - Precalculus with Limits

Question

Consider an initial deposit of \(P\) dollars in an account with an annual interest rate \(r\) , compounded monthly. At the end of each month, a withdrawal of \(W\) dollars will occur and the account will be depleted in \(t\) years. The amount of the initial deposit required is $$\begin{aligned} P=W\left(1+\frac{r}{12}\right)^{-1}+W\left(1+\frac{r}{12}\right)^{-2}+\cdots+& \\\ W\left(1+\frac{r}{12}\right)^{-12 t} & \end{aligned}$$ Show that the initial deposit is $$P=W\left(\frac{12}{r}\right)\left[1-\left(1+\frac{r}{12}\right)^{-12 t}\right]$$

Step-by-Step Solution

Answer

The initial deposit \( P \) can indeed be expressed as \( P = W\left(\dfrac{12}{r}\right)\left[1-\left(1 + \dfrac{r}{12}\right)^{-12t}\right] \)

Step 1: Understand the problem variables

First, let's understand and identify the terms. Here, \( P \) represents the initial deposit, \( W \) represents monthly withdrawal, \( r \) represents annual interest rate, and \( t \) is the time in years.

Step 2: Identify the common ratio

The common ratio for our geometric series is \( \left(1 + \dfrac{r}{12}\right)^{-1} \). This is the value each term is multiplied by to get the next term of the series.

Step 3: Use the sum formula of a geometric series

The sum \( S \) of \( n \) terms in a geometric series with first term \( a \) and common ratio \( r \) is: \( S = a \cdot \dfrac{1 - r^n}{1 - r} \). Comparing this to our expression, our first term \( a \) is \( W \) and \( n \) is \( 12t \)

Step 4: Substitute values into the sum formula:

Substitute \( n = 12t \), \( a = W \), and \( r = (1 + \dfrac{r}{12})^{-1} \) into the formula for \( S \): \( P = W \cdot \dfrac{1 - \left(1 + \dfrac{r}{12}\right)^{-12t}}{1 -(1 + \dfrac{r}{12})^{-1}} \)

Step 5: Simplify the expression

Simplifying the denominator by cross multiplication, we get \( P = W\left(\dfrac{12}{r}\right)\left[1-\left(1 + \dfrac{r}{12}\right)^{-12t}\right] \)