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Cálculo Activo

Matthew Boelkins

Section 8.2 Series Geométricas

Muchas secuencias importantes se generan por adición. En Preview Activity 8.2.1, vemos un ejemplo de una secuencia que está conectada a una suma.

Actividad Introductoria 8.2.1.

La warfarina es un anticoagulante que previene la coagulación de la sangre; a menudo se prescribe a las víctimas de un derrame cerebral para ayudar a asegurar el flujo sanguíneo. El nivel de warfarina tiene que alcanzar una cierta concentración en la sangre para ser efectivo.
Supón que la warfarina es tomada por un paciente en particular en una dosis de 5 mg cada día. El medicamento es absorbido por el cuerpo y algo es excretado del sistema entre dosis. Supón que al final de un período de 24 horas, el 8% del medicamento permanece en el cuerpo. Sea \(Q(n)\) la cantidad (en mg) de warfarina en el cuerpo antes de que se administre la \((n+1)\)ava dosis del medicamento.
  1. Explica por qué \(Q(1) = 5 \times 0.08\) mg.
  2. Explica por qué \(Q(2) = (5+Q(1)) \times 0.08\) mg. Luego muestra que
    \begin{equation*} Q(2) = (5 \times 0.08)\left(1+0.08\right) \text{mg}\text{.} \end{equation*}
  3. Explica por qué \(Q(3) = (5+Q(2)) \times 0.08\) mg. Luego muestra que
    \begin{equation*} Q(3) = (5 \times 0.08)\left(1+0.08+0.08^2\right) \text{mg}\text{.} \end{equation*}
  4. Explica por qué \(Q(4) = (5+Q(3)) \times 0.08\) mg. Luego muestra que
    \begin{equation*} Q(4) = (5 \times 0.08)\left(1+0.08+0.08^2+0.08^3\right) \text{mg}\text{.} \end{equation*}
  5. Hay un patrón que deberías ver emergiendo. Usa este patrón para encontrar una fórmula para \(Q(n)\text{,}\) donde \(n\) es un número entero positivo arbitrario.
  6. Completa Table 8.2.1 con valores de \(Q(n)\) para los valores de \(n\) proporcionados (reportando \(Q(n)\) con 10 decimales). ¿Qué parece estar ocurriendo con la secuencia \(Q(n)\) a medida que \(n\) aumenta?
    Table 8.2.1. Valores de \(Q(n)\) para valores seleccionados de \(n\)
    \(n\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\) \(7\) \(8\) \(9\) \(10\)
    \(Q(n)\) \(0.40\)                           

Subsection 8.2.1 Series Geométricas

En Preview Activity 8.2.1 encontramos la suma
\begin{equation*} (5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n-1}\right) \end{equation*}
para el nivel a largo plazo de Warfarina en el sistema del paciente. Esta suma tiene la forma
\begin{equation} a+ar+ar^2+ \cdots + ar^{n-1}\tag{8.2.1} \end{equation}
donde \(a=5 \times 0.08\) y \(r=0.08\text{.}\) Tal suma se llama una serie geométrica finita con razón \(r\text{.}\)

Activity 8.2.2.

Sean \(a\) y \(r\) números reales (con \(r \ne 1\)) y sea
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.} \end{equation*}
En esta actividad encontraremos una fórmula abreviada para \(S_n\) que no implique una suma de \(n\) términos.
  1. Multiplica \(S_n\) por \(r\text{.}\) ¿Cómo se ve la suma resultante?
  2. Resta \(rS_n\) de \(S_n\) y explica por qué
    \begin{equation} S_n - rS_n = a - ar^n\text{.}\tag{8.2.2} \end{equation}
  3. Resuelve la ecuación (8.2.2) para \(S_n\) para encontrar una fórmula simple para \(S_n\) que no implique sumar \(n\) términos.
La suma de los términos de una secuencia se llama una serie. Resumimos el resultado de Activity 8.2.2 de la siguiente manera.
Una serie geométrica finita \(S_n\) es una suma de la forma
\begin{equation} S_n = a + ar + ar^2 + \cdots + ar^{n-1}\text{,}\tag{8.2.3} \end{equation}
donde \(a\) y \(r\) son números reales tales que \(r \ne 1\text{.}\) La serie geométrica finita \(S_n\) se puede escribir más simplemente como
\begin{equation} S_n = a+ar+ar^2+ \cdots + ar^{n-1} = \frac{a(1-r^n)}{1-r}\text{.}\tag{8.2.4} \end{equation}
Ahora aplicamos la Ecuación (8.2.4) al ejemplo que involucra Warfarina de Preview Activity 8.2.1. Recuerda que
\begin{equation*} Q(n)=(5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n-1}\right) \text{mg}\text{,} \end{equation*}
así que \(Q(n)\) es una serie geométrica con \(a=5 \times 0.08 = 0.4\) y \(r = 0.08\text{.}\) Así,
\begin{equation*} Q(n) = 0.4\left(\frac{1-0.08^n}{1-0.08}\right) = \frac{1}{2.3} \left(1-0.08^n\right)\text{.} \end{equation*}
Nota que a medida que \(n\) tiende a infinito, el valor de \(0.08^n\) tiende a 0. Así,
\begin{equation*} \lim_{n \to \infty} Q(n) = \lim_{n \to \infty} \frac{1}{2.3} \left(1-0.08^n\right) = \frac{1}{2.3} \approx 0.435\text{.} \end{equation*}
Por lo tanto, el nivel a largo plazo de Warfarina en la sangre bajo estas condiciones es \(\frac{1}{2.3}\text{,}\) que es aproximadamente 0.435 mg.
Para determinar el efecto a largo plazo de la Warfarina, consideramos una serie geométrica finita de \(n\) términos, y luego consideramos qué pasaba cuando se permitía que \(n\) creciera sin límite. En este sentido, en realidad estábamos interesados en una serie geométrica infinita (el resultado de dejar que \(n\) tiende a infinito en la suma finita).

Definition 8.2.2.

Una serie geométrica infinita es una suma infinita de la forma
\begin{equation} a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{.}\tag{8.2.5} \end{equation}
El valor de \(r\) en la serie geométrica (8.2.5) se llama la razón común de la serie porque la razón del término (\(n+1\))-ésimo, \(ar^n\text{,}\) al término \(n\)-ésimo, \(ar^{n-1}\text{,}\) es siempre \(r\text{:}\)
\begin{equation*} \frac{ar^n}{ar^{n-1}} = r\text{.} \end{equation*}
Las series geométricas son comunes en matemáticas y surgen naturalmente en muchas situaciones diferentes. Como un ejemplo familiar, supón que queremos escribir el número con expansión decimal repetida
\begin{equation*} N=0.1212\overline{12} \end{equation*}
como un número racional. Observa que
\begin{align*} N \amp = 0.12 + 0.0012 + 0.000012 + \cdots\\ \amp = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.} \end{align*}
Esta es una serie geométrica infinita con \(a=\frac{12}{100}\) y \(r = \frac{1}{100}\text{.}\)
Usando la fórmula para el valor de una suma geométrica finita, también podemos desarrollar una fórmula para el valor de una serie geométrica infinita. Exploramos esta idea en la siguiente actividad.

Activity 8.2.3.

Sea \(r \ne 1\) y \(a\) números reales y sea
\begin{equation*} S = a+ar+ar^2 + \cdots ar^{n-1} + \cdots \end{equation*}
una serie geométrica infinita. Para cada número entero positivo \(n\text{,}\) sea
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.} \end{equation*}
Recuerda que
\begin{equation*} S_n = a\frac{1-r^n}{1-r}\text{.} \end{equation*}
  1. ¿A qué deberíamos permitir que \(n\) se acerque para que \(S_n\) se acerque a \(S\text{?}\)
  2. ¿Cuál es el valor de \(\lim_{n \to \infty} r^n\) para \(|r| \gt 1\text{?}\) ¿para \(|r| \lt 1\text{?}\) Explica.
  3. Si \(|r| \lt 1\text{,}\) usa la fórmula para \(S_n\) y tus observaciones en (a) y (b) para explicar por qué \(S\) es finito y encuentra una fórmula resultante para \(S\text{.}\)
Ahora podemos encontrar el valor de la serie geométrica
\begin{equation*} N = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.} \end{equation*}
Usando \(a = \frac{12}{100}\) y \(r = \frac{1}{100}\text{,}\) vemos que
\begin{equation*} N = \frac{12}{100} \left(\frac{1}{1-\frac{1}{100}}\right) = \frac{12}{100} \left(\frac{100}{99}\right) = \frac{4}{33}\text{.} \end{equation*}
La suma de un número finito de términos de una serie geométrica infinita a menudo se llama una suma parcial de la serie. Así,
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1} = \sum_{k=0}^{n-1} ar^k\text{.} \end{equation*}
se llama la \(n\)-ésima suma parcial de la serie \(\sum_{k=0}^{\infty} ar^k\text{.}\) Resumimos nuestro trabajo reciente con series geométricas de la siguiente manera.
  • Una serie geométrica infinita es una suma infinita de la forma
    \begin{equation} a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{,}\tag{8.2.6} \end{equation}
    donde \(a\) y \(r\) son números reales tales que \(r \ne 0\text{.}\)
  • La \(n\)-ésima suma parcial \(S_n\) de una serie geométrica infinita es
    \begin{equation*} S_n = a+ar+ar^2+ \cdots + ar^{n-1}\text{.} \end{equation*}
  • Si \(|r| \lt 1\text{,}\) entonces usando el hecho de que \(S_n = a\frac{1-r^n}{1-r}\text{,}\) se sigue que la suma \(S\) de la serie geométrica infinita (8.2.6) es
    \begin{equation*} S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r} \end{equation*}

Activity 8.2.4.

Las fórmulas que hemos derivado para una serie geométrica infinita y su suma parcial han asumido que comenzamos a indexar las sumas en \(n=0\text{.}\) Si en cambio tenemos una suma que no comienza en \(n=0\text{,}\) podemos factorizar términos comunes y usar las fórmulas establecidas. Este proceso se ilustra en los ejemplos de esta actividad.
  1. Considera la suma
    \begin{equation*} \sum_{k=1}^{\infty} (2)\left(\frac{1}{3}\right)^k = (2)\left(\frac{1}{3}\right) + (2)\left(\frac{1}{3}\right)^2 + (2)\left(\frac{1}{3}\right)^3 + \cdots\text{.} \end{equation*}
    Elimina el factor común de \((2)\left(\frac{1}{3}\right)\) de cada término y encuentra la suma de la serie.
  2. Ahora deja que \(a\) y \(r\) sean números reales con \(-1\lt r\lt 1\text{.}\) Considera la suma
    \begin{equation*} \sum_{k=3}^{\infty} ar^k = ar^3+ar^4+ar^5 + \cdots\text{.} \end{equation*}
    Elimina el factor común de \(ar^3\) de cada término y encuentra la suma de la serie.
  3. Finalmente, consideramos el caso más general. Deja que \(a\) y \(r\) sean números reales con \(-1\lt r\lt 1\text{,}\) deja que \(n\) sea un número entero positivo, y considera la suma
    \begin{equation*} \sum_{k=n}^{\infty} ar^k = ar^n+ar^{n+1}+ar^{n+2} + \cdots\text{.} \end{equation*}
    Elimina el factor común de \(ar^n\) de cada término para encontrar la suma de la serie.

Subsection 8.2.2 Resumen

  • Una serie geométrica infinita es una suma infinita de la forma
    \begin{equation*} \sum_{k=0}^{\infty} ar^k \end{equation*}
    donde \(a\) y \(r\) son números reales y \(r \neq 0\text{.}\)
  • La \(n\)-ésima suma parcial de la serie geométrica \(\sum_{k=0}^{\infty} ar^k\) es
    \begin{equation*} S_n = \sum_{k=0}^{n-1} ar^k\text{.} \end{equation*}
    Una fórmula para la \(n\)-ésima suma parcial de una serie geométrica es
    \begin{equation*} S_n = a \frac{1-r^n}{1-r}\text{.} \end{equation*}
    Si \(|r| \lt 1\text{,}\) la serie geométrica infinita \(\sum_{k=0}^{\infty} ar^k\) tiene la suma finita \(\frac{a}{1-r}\text{.}\)

Exercises 8.2.3 Exercises

1. Fourth term of a geometric sequence.

Find the \(4^{th}\) term of the geometric sequence
\(-1 , -3.5 , -12.25 , ...\)
Answer:

2. A geometric series.

Find the sum of the series
\(\displaystyle 2 + \frac{2}{7} + \frac{2}{49} + ... + \frac{2}{7^{n-1}} + ...\text{.}\)
Answer:

3. A series that is not geometric.

Determine the sum of the following series.
\begin{equation*} \sum_{n=1}^\infty \left(\frac{3^n + 8^n}{12 ^n}\right) \end{equation*}

4. Two sums of geometric sequences.

Find the sum of each of the geometric series given below. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation.
A. \(-15 + 5 - {5\over 3} + {5\over 9} - {5\over 27} + {5\over 81} - \cdots =\)
B. \(\sum\limits_{n=4}^{17} \left({1\over 2}\right)^n =\)

5.

There is an old question that is often used to introduce the power of geometric growth. Here is one version. Suppose you are hired for a one month (30 days, working every day) job and are given two options to be paid.
Option 1.
You can be paid $500 per day or
Option 2.
You can be paid 1 cent the first day, 2 cents the second day, 4 cents the third day, 8 cents the fourth day, and so on, doubling the amount you are paid each day.
  1. How much will you be paid for the job in total under Option 1?
  2. Complete Table 8.2.3 to determine the pay you will receive under Option 2 for the first 10 days.
    Table 8.2.3. Option 2 payments
    Day Pay on this day Total amount paid to date
    \(1\) \(\dollar0.01\) \(\dollar0.01\)
    \(2\) \(\dollar0.02\) \(\dollar0.03\)
    \(3\)
    \(4\)
    \(5\)
    \(6\)
    \(7\)
    \(8\)
    \(9\)
    \(10\)
  3. Find a formula for the amount paid on day \(n\text{,}\) as well as for the total amount paid by day \(n\text{.}\) Use this formula to determine which option (1 or 2) you should take.

6.

Suppose you drop a golf ball onto a hard surface from a height \(h\text{.}\) The collision with the ground causes the ball to lose energy and so it will not bounce back to its original height. The ball will then fall again to the ground, bounce back up, and continue. Assume that at each bounce the ball rises back to a height \(\frac{3}{4}\) of the height from which it dropped. Let \(h_n\) be the height of the ball on the \(n\)th bounce, with \(h_0 = h\text{.}\) In this exercise we will determine the distance traveled by the ball and the time it takes to travel that distance.
  1. Determine a formula for \(h_1\) in terms of \(h\text{.}\)
  2. Determine a formula for \(h_2\) in terms of \(h\text{.}\)
  3. Determine a formula for \(h_3\) in terms of \(h\text{.}\)
  4. Determine a formula for \(h_n\) in terms of \(h\text{.}\)
  5. Write an infinite series that represents the total distance traveled by the ball. Then determine the sum of this series.
  6. Next, let’s determine the total amount of time the ball is in the air.
    1. When the ball is dropped from a height \(H\text{,}\) if we assume the only force acting on it is the acceleration due to gravity, then the height of the ball at time \(t\) is given by
      \begin{equation*} H - \frac{1}{2}gt^2\text{.} \end{equation*}
      Use this formula to determine the time it takes for the ball to hit the ground after being dropped from height \(H\text{.}\)
    2. Use your work in the preceding item, along with that in (a)-(e) above to determine the total amount of time the ball is in the air.

7.

Suppose you play a game with a friend that involves rolling a standard six-sided die. Before a player can participate in the game, he or she must roll a six with the die. Assume that you roll first and that you and your friend take alternate rolls. In this exercise we will determine the probability that you roll the first six.
  1. Explain why the probability of rolling a six on any single roll (including your first turn) is \(\frac{1}{6}\text{.}\)
  2. If you don’t roll a six on your first turn, then in order for you to roll the first six on your second turn, both you and your friend had to fail to roll a six on your first turns, and then you had to succeed in rolling a six on your second turn. Explain why the probability of this event is
    \begin{equation*} \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)\text{.} \end{equation*}
  3. Now suppose you fail to roll the first six on your second turn. Explain why the probability is
    \begin{equation*} \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) \end{equation*}
    that you to roll the first six on your third turn.
  4. The probability of you rolling the first six is the probability that you roll the first six on your first turn plus the probability that you roll the first six on your second turn plus the probability that your roll the first six on your third turn, and so on. Explain why this probability is
    \begin{equation*} \frac{1}{6} + \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) + \cdots\text{.} \end{equation*}
    Find the sum of this series and determine the probability that you roll the first six.

8.

The goal of a federal government stimulus package is to positively affect the economy. Economists and politicians quote numbers like “\(k\) million jobs and a net stimulus to the economy of \(n\) billion of dollars.” Where do they get these numbers? Let’s consider one aspect of a stimulus package: tax cuts. Economists understand that tax cuts or rebates can result in long-term spending that is many times the amount of the rebate. For example, assume that for a typical person, 75% of her entire income is spent (that is, put back into the economy). Further, assume the government provides a tax cut or rebate that totals \(P\) dollars for each person.
  1. The tax cut of \(P\) dollars is income for its recipient. How much of this tax cut will be spent?
  2. In this simple model, we will say that the spent portion of the tax cut/rebate from part (a) then becomes income for another person who, in turn, spends 75% of this income. After this ``second round" of spent income, how many total dollars have been added to the economy as a result of the original tax cut/rebate?
  3. This second round of spending becomes income for another group who spend 75% of this income, and so on. In economics this is called the multiplier effect. Explain why an original tax cut/rebate of \(P\) dollars will result in multiplied spending of
    \begin{equation*} 0.75P(1+0.75+0.75^2+ \cdots )\text{.} \end{equation*}
    dollars.
  4. Based on these assumptions, how much stimulus will a 200 billion dollar tax cut/rebate to consumers add to the economy, assuming consumer spending remains consistent forever.

9.

Like stimulus packages, home mortgages and foreclosures also impact the economy. A problem for many borrowers is the adjustable rate mortgage, in which the interest rate can change (and usually increases) over the duration of the loan, causing the monthly payments to increase beyond the ability of the borrower to pay. Most financial analysts recommend fixed rate loans, ones for which the monthly payments remain constant throughout the term of the loan. In this exercise we will analyze fixed rate loans.
When most people buy a large ticket item like car or a house, they have to take out a loan to make the purchase. The loan is paid back in monthly installments until the entire amount of the loan, plus interest, is paid. With a loan, we borrow money, say \(P\) dollars (called the principal), and pay off the loan at an interest rate of \(r\)%. To pay back the loan we make regular monthly payments, some of which goes to pay off the principal and some of which is charged as interest. In most cases, the interest is computed based on the amount of principal that remains at the beginning of the month. We assume a fixed rate loan, that is one in which we make a constant monthly payment \(M\) on our loan, beginning in the original month of the loan.
Suppose you want to buy a house. You have a certain amount of money saved to make a down payment, and you will borrow the rest to pay for the house. Of course, for the privilege of loaning you the money, the bank will charge you interest on this loan, so the amount you pay back to the bank is more than the amount you borrow. In fact, the amount you ultimately pay depends on three things: the amount you borrow (called the principal), the interest rate, and the length of time you have to pay off the loan plus interest (called the duration of the loan). For this example, we assume that the interest rate is fixed at \(r\)%.
To pay off the loan, each month you make a payment of the same amount (called installments). Suppose we borrow \(P\) dollars (our principal) and pay off the loan at an interest rate of \(r\)% with regular monthly installment payments of \(M\) dollars. So in month 1 of the loan, before we make any payments, our principal is \(P\) dollars. Our goal in this exercise is to find a formula that relates these three parameters to the time duration of the loan.
We are charged interest every month at an annual rate of \(r\)%, so each month we pay \(\frac{r}{12}\)% interest on the principal that remains. Given that the original principal is \(P\) dollars, we will pay \(\left(\frac{0.0r}{12}\right)P\) dollars in interest on our first payment. Since we paid \(M\) dollars in total for our first payment, the remainder of the payment (\(M-\left(\frac{r}{12}\right)P\)) goes to pay down the principal. So the principal remaining after the first payment (let’s call it \(P_1\)) is the original principal minus what we paid on the principal, or
\begin{equation*} P_1 = P - \left( M - \left(\frac{r}{12}\right)P\right) = \left(1 + \frac{r}{12}\right)P - M\text{.} \end{equation*}
As long as \(P_1\) is positive, we still have to keep making payments to pay off the loan.
  1. Recall that the amount of interest we pay each time depends on the principal that remains. How much interest, in terms of \(P_1\) and \(r\text{,}\) do we pay in the second installment?
  2. How much of our second monthly installment goes to pay off the principal? What is the principal \(P_2\text{,}\) or the balance of the loan, that we still have to pay off after making the second installment of the loan? Write your response in the form \(P_2 = ( \ )P_1 - ( \ )M\text{,}\) where you fill in the parentheses.
  3. Show that \(P_2 = \left(1 + \frac{r}{12}\right)^2P - \left[1 + \left(1+\frac{r}{12}\right)\right] M\text{.}\)
  4. Let \(P_3\) be the amount of principal that remains after the third installment. Show that
    \begin{equation*} P_3 = \left(1 + \frac{r}{12}\right)^3P - \left[1 + \left(1+\frac{r}{12}\right) + \left(1+\frac{r}{12}\right)^2 \right] M\text{.} \end{equation*}
  5. If we continue in the manner described in the problems above, then the remaining principal of our loan after \(n\) installments is
    \begin{equation} P_n = \left(1 + \frac{r}{12}\right)^nP - \left[\displaystyle \sum_{k=0}^{n-1} \left(1+\frac{r}{12}\right)^k \right] M\text{.}\tag{8.2.7} \end{equation}
    This is a rather complicated formula and one that is difficult to use. However, we can simplify the sum if we recognize part of it as a partial sum of a geometric series. Find a formula for the sum
    \begin{equation} \displaystyle \sum_{k=0}^{n-1} \left(1+\frac{r}{12}\right)^k\text{.}\tag{8.2.8} \end{equation}
    and then a general formula for \(P_n\) that does not involve a sum.
  6. It is usually more convenient to write our formula for \(P_n\) in terms of years rather than months. Show that \(P(t)\text{,}\) the principal remaining after \(t\) years, can be written as
    \begin{equation} P(t) = \left(P - \frac{12M}{r}\right)\left(1+\frac{r}{12}\right)^{12t} + \frac{12M}{r}\text{.}\tag{8.2.9} \end{equation}
  7. Now that we have analyzed the general loan situation, we apply formula (8.2.9) to an actual loan. Suppose we charge $1,000 on a credit card for holiday expenses. If our credit card charges 20% interest and we pay only the minimum payment of $25 each month, how long will it take us to pay off the $1,000 charge? How much in total will we have paid on this $1,000 charge? How much total interest will we pay on this loan?
  8. Now we consider larger loans, e.g., automobile loans or mortgages, in which we borrow a specified amount of money over a specified period of time. In this situation, we need to determine the amount of the monthly payment we need to make to pay off the loan in the specified amount of time. In this situation, we need to find the monthly payment \(M\) that will take our outstanding principal to \(0\) in the specified amount of time. To do so, we want to know the value of \(M\) that makes \(P(t) = 0\) in formula (8.2.9). If we set \(P(t) = 0\) and solve for \(M\text{,}\) it follows that
    \begin{equation*} M = \frac{rP \left(1+\frac{r}{12}\right)^{12t}}{12\left(\left(1+\frac{r}{12}\right)^{12t} - 1 \right)}\text{.} \end{equation*}
    1. Suppose we want to borrow $15,000 to buy a car. We take out a 5 year loan at 6.25%. What will our monthly payments be? How much in total will we have paid for this $15,000 car? How much total interest will we pay on this loan?
    2. Suppose you charge your books for winter semester on your credit card. The total charge comes to $525. If your credit card has an interest rate of 18% and you pay $20 per month on the card, how long will it take before you pay off this debt? How much total interest will you pay?
    3. Say you need to borrow $100,000 to buy a house. You have several options on the loan:
      • 30 years at 6.5%
      • 25 years at 7.5%
      • 15 years at 8.25%.
      1. What are the monthly payments for each loan?
      2. Which mortgage is ultimately the best deal (assuming you can afford the monthly payments)? In other words, for which loan do you pay the least amount of total interest?
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