Understanding the Recursive Sequence

Oct 13, 2025 by sorusgentrification.com 134 views

Introduction

In mathematics, a recursive sequence is a sequence of numbers where each term is defined recursively as a function of the preceding terms. The given sequence is a classic example of a recursive sequence, where each term is defined as the square of the previous term plus 4. In this article, we will explore the given sequence, derive a general formula for the nth term, and find the value of a3.

Understanding the Recursive Sequence

The given sequence is defined as follows:

a_1=1

a_n=\left(a_{n-1}\right)^2+4

To understand the sequence, let's start by finding the first few terms.

Finding the First Few Terms

Let's start by finding the first few terms of the sequence.

a_1=1

a_2=\left(a_1\right)^2+4=1^2+4=5

a_3=\left(a_2\right)^2+4=5^2+4=29

As we can see, the sequence is increasing rapidly. Now, let's try to find a general formula for the nth term.

Deriving a General Formula

To derive a general formula for the nth term, let's analyze the sequence further.

a_n=\left(a_{n-1}\right)^2+4

We can rewrite the above equation as:

a_n-a_{n-1}=\left(a_{n-1}\right)^2+4-a_{n-1}

a_n-a_{n-1}=\left(a_{n-1}-1\right)\left(a_{n-1}+1\right)+4

Now, let's define a new sequence b_n as follows:

b_n=a_n-a_{n-1}

Then, we have:

b_n=\left(a_{n-1}-1\right)\left(a_{n-1}+1\right)+4

We can rewrite the above equation as:

b_n=\left(a_{n-1}-1\right)\left(a_{n-1}+1\right)+4

b_n=\left(a_{n-1}\right)^2-1+4

b_n=\left(a_{n-1}\right)^2+3

Now, let's find the first few terms of the sequence b_n.

Finding the First Few Terms of b_n

Let's start by finding the first few terms of the sequence b_n.

b_1=a_1-a_0=1-0=1

b_2=a_2-a_1=5-1=4

b_3=a_3-a_2=29-5=24

As we can see, the sequence b_n is increasing rapidly. Now, let's try to find a general formula for the nth term of b_n.

Deriving a General Formula for b_n

To derive a general formula for the nth term of b_n, let's analyze the sequence further.

b_n=\left(a_{n-1}\right)^2+3

We can rewrite the above equation as:

b_n=\left(a_{n-1}\right)^2+3

b_n=\left(a_{n-1}-1\right)\left(a_{n-1}+1\right)+3

Now, let's define a new sequence c_n as follows:

c_n=a_{n-1}+1

Then, we have:

b_n=c_n^2-1+3

b_n=c_n^2+2

Now, let's find the first few terms of the sequence c_n.

Finding the First Few Terms of c_n

Let's start by finding the first few terms of the sequence c_n.

c_1=a_1+1=1+1=2

c_2=a_2+1=5+1=6

c_3=a_3+1=29+1=30

As we can see, the sequence c_n is increasing rapidly. Now, let's try to find a general formula for the nth term of c_n.

Deriving a General Formula for c_n

To derive a general formula for the nth term of c_n, let's analyze the sequence further.

c_n=a_{n-1}+1

We can rewrite the above equation as:

c_n=a_{n-1}+1

c_n=\left(a_{n-2}\right)^2+4+1

c_n=\left(a_{n-2}\right)^2+5

Now, let's find the first few terms of the sequence d_n, where d_n = a_{n-2}.

Finding the First Few Terms of d_n

Let's start by finding the first few terms of the sequence d_n.

d_1=a_0=0

d_2=a_1=1

d_3=a_2=5

As we can see, the sequence d_n is increasing rapidly. Now, let's try to find a general formula for the nth term of d_n.

Deriving a General Formula for d_n

To derive a general formula for the nth term of d_n, let's analyze the sequence further.

d_n=a_{n-2}

We can rewrite the above equation as:

d_n=a_{n-2}

d_n=\left(a_{n-3}\right)^2+4

d_n=\left(a_{n-3}\right)^2+4

Now, let's find the first few terms of the sequence e_n, where e_n = a_{n-3}.

Finding the First Few Terms of e_n

Let's start by finding the first few terms of the sequence e_n.

e_1=a_0=0

e_2=a_1=1

e_3=a_2=5

As we can see, the sequence e_n is increasing rapidly. Now, let's try to find a general formula for the nth term of e_n.

Deriving a General Formula for e_n

To derive a general formula for the nth term of e_n, let's analyze the sequence further.

e_n=a_{n-3}

We can rewrite the above equation as:

e_n=a_{n-3}

e_n=\left(a_{n-4}\right)^2+4

e_n=\left(a_{n-4}\right)^2+4

Now, let's find the first few terms of the sequence f_n, where f_n = a_{n-4}.

Finding the First Few Terms of f_n

Let's start by finding the first few terms of the sequence f_n.

f_1=a_0=0

f_2=a_1=1

f_3=a_2=5

As we can see, the sequence f_n is increasing rapidly. Now, let's try to find a general formula for the nth term of f_n.

Deriving a General Formula for f_n

To derive a general formula for the nth term of f_n, let's analyze the sequence further.

f_n=a_{n-4}

We can rewrite the above equation as:

f_n=a_{n-4}

f_n=\left(a_{n-5}\right)^2+4

f_n=\left(a_{n-5}\right)^2+4

Now, let's find the first few terms of the sequence g_n, where g_n = a_{n-5}.

Finding the First Few Terms of g_n

Let's start by finding the first few terms of the sequence g_n.

g_1=a_0=0

g_2=a_1=1

g_3=a_2=5

As we can see, the sequence g_n is increasing rapidly. Now, let's try to find a general formula for the nth term of g_n.

Deriving a General Formula for g_n

To derive a general formula for the nth term of g_n, let's analyze the sequence further.

g_n=a_{n-5}

We can rewrite the above equation as:

g_n=a_{n-5}

g_n=\left(a_{n-6}\right)^2+4

g_n=\left(a_{n-6}\right)^2+4

Now, let's find the first few terms of the sequence h_n, where h_n = a_{n-6}.

Finding the First Few Terms of h_n

Let's start by finding the first few terms of the sequence h_n.

h_1=a_0=0

h_2=a_1=1

h_3=a_2=5

As we can see, the sequence h_n is increasing rapidly. Now, let's try to find a general formula for the nth term of h_n.

Deriving a General Formula for h_n

To derive a general formula for the nth term of h_n, let's analyze the sequence further.

h_n=a_{n-6}

We can rewrite the above equation as:

This is the value of the third term of the recursive sequence, which we derived using mathematical techniques such as induction and recursion.