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Chapter 4: Problem 22

Compound Interest. Assume that \(R\) dollars is deposited in an account that has an interest rate of \(i\) for each compounding period. A compounding period is some specified time period such as a month or a year. For each integer \(n\) with \(n \geq 0,\) let \(V_{n}\) be the amount of money in an account at the end of the \(n\) th compounding period. Then $$ \begin{array}{rlrl} V_{1}=R+i \cdot R & V_{2} & =V_{1}+i \cdot V_{1} \\ =R(1+i) & & =(1+i) V_{1} \\ & = & (1+i)^{2} R . \end{array} $$ (a) Explain why \(V_{3}=V_{2}+i \cdot V_{2}\). Then use the formula for \(V_{2}\) to determine a formula for \(V_{3}\) in terms of \(i\) and \(R\). (b) Determine a recurrence relation for \(V_{n+1}\) in terms of \(i\) and \(V_{n}\). (c) Write the recurrence relation in Part ( \(22 \mathrm{~b}\) ) so that it is in the form of a recurrence relation for a geometric sequence. What is the initial term of the geometric sequence and what is the common ratio? (d) Use Proposition 4.14 to determine a formula for \(V_{n}\) in terms of \(I, R\), and \(n\).

Short Answer

Expert verified
(a) \(V_3 = (1+i)^3R\) (b) \(V_{n+1} = V_n + i\cdot V_n\) (c) Initial term: \(V_1 = R(1+i)\), Common ratio: \((1+i)\) (d) \(V_n = R(1+i)^n\)

Step by step solution

01

(a) Understanding \(V_3\) and deriving a formula for it

We are given that \(V_1 = R + i\cdot R\) and \(V_2 = (1+i)^2R\). To find a formula for \(V_3\), we first need to understand the relation \(V_3 = V_2 + i\cdot V_2\). This relation states that the amount at the end of the third period, \(V_3\), is equal to the amount at the end of the second period, \(V_2\), plus the interest earned on \(V_2\) during the third period, which is \(i\cdot V_2\). This relationship makes sense, as the amount in the account increases by the interest earned on the previous amount. Now we can use the formula for \(V_2\) to find a formula for \(V_3\): \[V_3 = V_2 + i\cdot V_2 = (1+i)^2R + i(1+i)^2R = (1+i)^3R\]
02

(b) Recurrence relation for \(V_{n+1}\)

To find a recurrence relation for \(V_{n+1}\), we need to extend the pattern we found for \(V_2\) and \(V_3\). As we've seen before, the interest earned on the previous amount is added to the amount at the end of each compounding period. So, for \(V_{n+1}\) in terms of \(i\) and \(V_n\), we can say: \[V_{n+1} = V_n + i\cdot V_n\]
03

(c) Recurrence relation as a geometric sequence

To rewrite the recurrence relation in the form of a geometric sequence, we can factor out \(V_n\) from the expression: \[V_{n+1} = V_n(1+i)\] This is a geometric sequence with an initial term \(V_1 = R(1+i)\) and a common ratio of \((1+i)\).
04

(d) Formula for \(V_n\) using Proposition 4.14

Finally, we can use Proposition 4.14, which states that the formula for the nth term of a geometric sequence is given by: \[A_n = A_1\cdot r^{n-1}\] In our case, \(A_n = V_n\), \(A_1 = V_1 = R(1+i)\), and the common ratio \(r = (1+i)\). So, we can find a formula for \(V_n\) as follows: \[V_n = V_1\cdot r^{n-1} = R(1+i)((1+i)^{n-1}) = R(1+i)^n\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recurrence Relation
A recurrence relation is a mathematical expression that defines each term of a sequence based on its predecessors. In our context of compound interest, this implies that the amount of money available at the end of each compounding period depends on the amount from the previous period. The recurrence relation can be seen as a formula that shows how one step in a sequence is generated from the preceding step(s).
  • For compound interest, this relationship is expressed as \( V_{n+1} = V_n + i \cdot V_n \), where \( V_{n} \) is the current amount and \( V_{n+1} \) is the amount after the next compounding period.
  • This concept helps us understand how the interest earned during each period is added back to the existing amount to accumulate over time.
This simple relation is foundational for working with sequences, as it allows prediction of future values based on the present state.
Geometric Sequence
A geometric sequence is a sequence in which each term is derived by multiplying the previous term by a constant factor known as the common ratio. In the case of compound interest, this constant factor is represented by the expression \( 1+i \).
  • The sequence formed by the compound interest model is geometric because each term grows by the same ratio \((1+i)\) relative to the term before it.
  • The initial term of this sequence is \( V_1 = R(1+i) \).
  • The sequence can be represented as \( V_1, V_2, V_3, \ldots \) where each term \( V_n \) equals \( V_1(1+i)^{n-1} \).
Understanding geometric sequences is key to grasping how investments grow exponentially over time with a fixed interest rate.
Compounding Period
The concept of a compounding period is central to understanding how compound interest accumulates. A compounding period is a specified time frame after which interest is calculated and added to the principal amount.
  • Common compounding periods include monthly, quarterly, semi-annually, and annually.
  • The more frequent the compounding periods within a year, the more often interest is calculated and added, leading to potentially greater overall returns.
  • The formula \( V_n = R(1+i)^n \) takes the number of compounding periods into account, where \( i \) is the interest rate per period, not per year.
Understanding how compounding periods affect growth can help you make better financial decisions regarding investments and savings.
Interest Rate
The interest rate is a critical factor in calculating compound interest as it determines how much additional money will be added during each compounding period. An interest rate generally represents the annual rate, but in compound interest calculations, it is the rate per compounding period.
  • When calculating interest for each compounding period, the annual rate must be adjusted to fit the period. For example, if compounding monthly, the interest rate \(i\) would be \(\frac{\text{annual rate}}{12}\).
  • The formula \( V_n = R(1+i)^n \) shows how the interest rate \(i\) influences the growth of the principal \(R\) over time \(n\).
  • Higher interest rates or more frequent compounding periods (monthly vs. annually) result in more compound interest earnings.
Understanding interest rates and their impact on compound growth is essential for making informed financial choices, particularly in long-term investments.

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Most popular questions from this chapter

Most of the work done in constructing a proof by induction is usually in proving the inductive step. This was certainly the case in Proposition 4.2 . However, the basis step is an essential part of the proof. Without it, the proof is incomplete. To see this, let \(P(n)\) be $$ 1+2+\cdots+n=\frac{n^{2}+n+1}{2} $$ (a) Let \(k \in \mathbb{N}\). Complete the following proof that if \(P(k)\) is true, then \(P(k+1)\) is true. Let \(k \in \mathbb{N}\). Assume that \(P(k)\) is true. That is, assume that $$ 1+2+\cdots+k=\frac{k^{2}+k+1}{2} $$ The goal is to prove that \(P(k+1)\) is true. That is, we need to prove that $$ 1+2+\cdots+k+(k+1)=\frac{(k+1)^{2}+(k+1)+1}{2} $$ To do this, we add \((k+1)\) to both sides of equation (1). This gives $$ \begin{aligned} 1+2+\cdots+k+(k+1) &=\frac{k^{2}+k+1}{2}+(k+1) \\ &=\cdots \end{aligned} $$ (b) Is \(P(1)\) true? Is \(P(2)\) true? What about \(P(3)\) and \(P(4) ?\) Explain how this shows that the basis step is an essential part of a proof by induction.

For which natural numbers \(n\) is \(n^{2}<2^{n} ?\) Justify your conclusion.

Which of the following sets are inductive sets? Explain. (a) \(\mathbb{Z}\) (c) \(\\{x \in \mathbb{Z} \mid x \leq 10\\}\) (b) \(\\{x \in \mathbb{N} \mid x \geq 4\\}\) (d) \(\\{1,2,3, \ldots, 500\\}\)

Is the following proposition true or false? Justify your conclusion. For each nonnegative integer \(n, 8^{n} \mid(4 n) !\)

(a) For each natural number \(n, 1+4+7+\cdots+(3 n-2)=\frac{n(3 n-1)}{2}\). So we let \(P(n)\) be the open sentence $$ 1+4+7+\cdots+(3 n-2) $$ Using \(n=1,\) we see that \(3 n-2=1\) and hence, \(P(1)\) is true. We now assume that \(P(k)\) is true. That is, $$ 1+4+7+\cdots+(3 k-2)=\frac{k(3 k-1)}{2} $$ We then see that $$ \begin{aligned} 1+4+7+\cdots+(3 k-2)+(3(k+1)-2) &=\frac{(k+1)(3 k+2)}{2} \\ \frac{k(3 k-1)}{2}+(3 k+1) &=\frac{(k+1)(3 k+2)}{2} \\ \frac{\left(3 k^{2}-k\right)+(6 k+2)}{2} &=\frac{3 k^{2}+5 k+2}{2} \\ \frac{3 k^{2}+5 k+2}{2} &=\frac{3 k^{2}+5 k+2}{2} . \end{aligned} $$ We have thus proved that \(P(k+1)\) is true, and hence, we have proved the proposition. (b) For each natural number \(n, 1+4+7+\cdots+(3 n-2)=\frac{n(3 n-1)}{2}\). Proof. We will prove this proposition using mathematical induction. So we let $$ P(n)=1+4+7+\cdots+(3 n-2) $$Using \(n=1,\) we see that \(P(1)=1\) and hence, \(P(1)\) is true. We now assume that \(P(k)\) is true. That is, $$ 1+4+7+\cdots+(3 k-2)=\frac{k(3 k-1)}{2} $$ We then see that $$ \begin{aligned} P(k+1) &=1+4+7+\cdots+(3 k-2)+(3(k+1)-2) \\ &=\frac{k(3 k-1)}{2}+3(k+1)-2 \\ &=\frac{3 k^{2}-k+6 k+6-4}{2} \\ &=\frac{3 k^{2}+5 k+2}{2} \\ &=\frac{(k+1)(3 k+2)}{2} \end{aligned} $$ We have thus proved that \(P(k+1)\) is true, and hence, we have proved the proposition. (c) All dogs are the same breed. Proof. We will prove this proposition using mathematical induction. For each natural number \(n,\) we let \(P(n)\) be Any set of \(n\) dogs consists entirely of dogs of the same breed. We will prove that for each natural number \(n, P(n)\) is true, which will prove that all dogs are the same breed. A set with only one dog consists entirely of dogs of the same breed and, hence, \(P(1)\) is true. So we let \(k\) be a natural number and assume that \(P(k)\) is true, that is, that every set of \(k\) dogs consists of dogs of the same breed. Now consider a set \(D\) of \(k+1\) dogs, where $$ D=\left\\{d_{1}, d_{2}, \ldots, d_{k}, d_{k+1}\right\\} $$ If we remove the \(\operatorname{dog} d_{1}\) from the set \(D\), we then have a set \(D_{1}\) of \(k\) dogs, and using the assumption that \(P(k)\) is true, these dogs must all be of the same breed. Similarly, if we remove \(d_{k+1}\) from the set \(D,\) we again have a set \(D_{2}\) of \(k\) dogs, and these dogs must all be of the same breed. Since \(D=D_{1} \cup D_{2},\) we have proved that all of the dogs in \(D\) must be of the same breed. This proves that if \(P(k)\) is true, then \(P(k+1)\) is true and, hence, by mathematical induction, we have proved that for each natural number \(n,\) any set of \(n\) dogs consists entirely of dogs of the same breed.
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