The Type aₙ₊₁ = qaₙ

This type of recurrence relation corresponds to a geometric sequence and embodies the characteristics of exponential growth.

The Recurrence Relation

Meaning: each term is $q$ times the previous term.

Solution Method

Starting from the recurrence relation:

$$\begin{aligned} a_2 &= qa_1 \\ a_3 &= qa_2 = q^2 a_1 \\ a_4 &= qa_3 = q^3 a_1 \\ &\vdots \\ a_n &= q^{n-1} a_1 \end{aligned}$$

This is exactly the general term formula of a geometric sequence!

Worked Examples

Example 1: Basic Solution

The sequence $\{a_n\}$ satisfies $a_1 = 2$ and $a_{n+1} = 3a_n$. Find $a_n$.

Solution:

This is of the type $a_{n+1} = qa_n$, where $q = 3$.

$a_n = a_1 \cdot q^{n-1} = 2 \cdot 3^{n-1}$

Example 2: Bacterial Reproduction

A certain bacterium divides once every hour, splitting into 2 each time. Initially there is 1 bacterium. Find the number of bacteria after $n$ hours.

Solution:

Let the number of bacteria after $n$ hours be $a_n$. Then:

• $a_1 = 1$
• $a_{n+1} = 2a_n$

This is a geometric-type recurrence with $q = 2$:

$a_n = 1 \cdot 2^{n-1} = 2^{n-1}$

Example 3: Compound Interest Calculation

Deposit 10000 yuan in a bank at an annual interest rate of 5% with compound interest. What is the amount after $n$ years?

Solution:

Let the amount after $n$ years be $a_n$ yuan. Then:

• $a_1 = 10000$
• $a_{n+1} = a_n \times (1 + 5\%) = 1.05a_n$

This is a geometric-type recurrence with $q = 1.05$:

$a_n = 10000 \cdot 1.05^{n-1}$

Practice Problems

Exercise 1

The sequence $\{a_n\}$ satisfies $a_1 = 3$ and $a_{n+1} = 2a_n$. Find $a_6$.

Exercise 2

The sequence $\{a_n\}$ satisfies $a_1 = 8$ and $a_{n+1} = \frac{1}{2}a_n$. Find the general term formula.

Exercise 3

The sequence $\{a_n\}$ satisfies $a_3 = 12$ and $a_{n+1} = 3a_n$. Find $a_1$ and the general term formula.

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Summary

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