Math Problem Statement

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]

Solution

(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)=P0(1+r)tP(t) = P_0 \cdot (1 + r)^t

where:

  • P(t)P(t) = population after tt years
  • P0P_0 = initial population
  • rr = growth rate (expressed as a decimal)
  • tt = time in years

In this case:

  • P0=50P_0 = 50 million
  • r=0.03r = 0.03 (3%)
  • t=25t = 25

Substitute the values into the formula:

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

Using a calculator, (1.03)252.09377(1.03)^{25} \approx 2.09377:

P(25)=502.09377104.6885 millionP(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 ii, compounded mm times per year:

Pk=[1+im]Pk1P_k = \left[ 1 + \frac{i}{m} \right] P_{k-1}

with initial condition P0=PP_0 = P (the initial amount deposited).

To find an explicit formula for PnP_n, let’s solve this recurrence relation:

  1. Recognize the pattern:

    For k=1k = 1: P1=[1+im]P0P_1 = \left[ 1 + \frac{i}{m} \right] P_0

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

    For k=3k = 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 nn: Pn=[1+im]nP0P_n = \left[ 1 + \frac{i}{m} \right]^n P_0

  2. Final formula:

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

This formula provides the amount of money after nn compounding periods, where PP is the initial deposit, ii is the annual interest rate, and mm 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.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Compound Interest
Recurrence Relations

Formulas

Exponential Growth Formula: P(t) = P_0 * (1 + r)^t
Compound Interest Formula: P_n = P * [1 + i/m]^n

Theorems

Exponential Growth Theorem
Compound Interest Theorem

Suitable Grade Level

Grades 10-12