Suppose the population of a country increases at a steady rate of 3% per year. If the

population is 50 million at a certain time, what will it be 25 years later? Provide a

non-recursive formula[10 marks]

e. As shown in Example 5.6.8, if a bank pays interest at a rate of i compounded m times

a year, then the amount of money Pk at the end of k time periods (where one time

period = 1/mth of a year) satisfies the recurrence relation Pk = [1+ 1 (i/m)] Pk-1 with

initial condition P0 = the initial amount deposited. Find an explicit formula for Pn. [10

marks]

### (a) **Population Growth Problem**

To find the population of a country that increases at a steady rate of 3% per year, starting from 50 million, we use the formula for exponential growth:

\[
P(t) = P_0 \cdot (1 + r)^t
\]

where:
- \( P(t) \) = population after \( t \) years
- \( P_0 \) = initial population
- \( r \) = growth rate (expressed as a decimal)
- \( t \) = time in years

In this case:
- \( P_0 = 50 \) million
- \( r = 0.03 \) (3%)
- \( t = 25 \)

Substitute the values into the formula:

\[
P(25) = 50 \cdot (1 + 0.03)^{25}
\]
\[
P(25) = 50 \cdot (1.03)^{25}
\]

Using a calculator, \((1.03)^{25} \approx 2.09377\):

\[
P(25) = 50 \cdot 2.09377 \approx 104.6885 \text{ million}
\]

So, the population 25 years later will be approximately **104.69 million**.

### (b) **Compound Interest Problem**

We are given a recurrence relation for the amount of money in an account that pays interest at a rate of \( i \), compounded \( m \) times per year:

\[
P_k = \left[ 1 + \frac{i}{m} \right] P_{k-1}
\]

with initial condition \( P_0 = P \) (the initial amount deposited).

To find an explicit formula for \( P_n \), let’s solve this recurrence relation:

1. **Recognize the pattern:**

   For \( k = 1 \):
   \[
   P_1 = \left[ 1 + \frac{i}{m} \right] P_0
   \]

   For \( k = 2 \):
   \[
   P_2 = \left[ 1 + \frac{i}{m} \right] P_1 = \left[ 1 + \frac{i}{m} \right]^2 P_0
   \]

   For \( k = 3 \):
   \[
   P_3 = \left[ 1 + \frac{i}{m} \right] P_2 = \left[ 1 + \frac{i}{m} \right]^3 P_0
   \]

   Continuing this pattern, for any \( n \):
   \[
   P_n = \left[ 1 + \frac{i}{m} \right]^n P_0
   \]

2. **Final formula:**

   \[
   P_n = P \cdot \left[ 1 + \frac{i}{m} \right]^n
   \]

This formula provides the amount of money after \( n \) compounding periods, where \( P \) is the initial deposit, \( i \) is the annual interest rate, and \( m \) is the number of times the interest is compounded per year.

---

### Do you want more details on these explanations or have any questions?

### 5 Related Questions:

1. How does the compound interest formula change if the compounding is continuous?
2. What would the population be after 50 years if the growth rate remains the same?
3. How can you derive the formula for exponential growth from differential equations?
4. If a bank offers different compounding frequencies (e.g., monthly, quarterly), how does that affect the final amount?
5. What is the difference between simple interest and compound interest calculations?

### Tip:
Always convert percentage rates into decimals when using them in formulas to avoid errors in calculations. For example, 3% becomes 0.03 when substituted into formulas.

Suppose the population of a country increases at a steady rate of 3% per year. If the population is 50 million at a certain time, what will it be 25 years later? Provide a non-recursive formula. Additionally, find an explicit formula for the compound interest recurrence relation P_k = [1 + i/m]P_(k-1) with initial condition P_0 (initial amount deposited).

Explore how to calculate population growth and compound interest using the exponential growth formula and compound interest formula. Understand how a steady growth rate of 3% affects population over 25 years, and find an explicit formula for compound interest based on a recurrence relation. Suitable for Grades 10-12.

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