Suppose a^n = -4an^(n-1) + 5a^(n-2) - 4a(n-3) and a^4 = -98, a^5 = 452, and a^6 = -2314. Find a^1,a^2,a^3

We are given the recurrence relation for \( a^n \) as follows:

\[
a^n = -4a^{n-1} + 5a^{n-2} - 4a^{n-3}
\]

Also, we are provided with:

\[
a^4 = -98, \quad a^5 = 452, \quad a^6 = -2314
\]

Our goal is to find \( a^1 \), \( a^2 \), and \( a^3 \).

### Step 1: Express the recurrence relation for specific values of \( n \)

We will use the recurrence relation for \( n = 6 \), \( n = 5 \), and \( n = 4 \).

#### Case \( n = 6 \):

\[
a^6 = -4a^5 + 5a^4 - 4a^3
\]

Substitute \( a^6 = -2314 \), \( a^5 = 452 \), and \( a^4 = -98 \):

\[
-2314 = -4(452) + 5(-98) - 4a^3
\]

Simplify the terms:

\[
-2314 = -1808 - 490 - 4a^3
\]

\[
-2314 = -2298 - 4a^3
\]

\[
-2314 + 2298 = -4a^3
\]

\[
-16 = -4a^3
\]

\[
a^3 = 4
\]

### Step 2: Use the recurrence relation for \( n = 5 \)

\[
a^5 = -4a^4 + 5a^3 - 4a^2
\]

Substitute \( a^5 = 452 \), \( a^4 = -98 \), and \( a^3 = 4 \):

\[
452 = -4(-98) + 5(4) - 4a^2
\]

Simplify the terms:

\[
452 = 392 + 20 - 4a^2
\]

\[
452 = 412 - 4a^2
\]

\[
452 - 412 = -4a^2
\]

\[
40 = -4a^2
\]

\[
a^2 = -10
\]

### Step 3: Use the recurrence relation for \( n = 4 \)

\[
a^4 = -4a^3 + 5a^2 - 4a^1
\]

Substitute \( a^4 = -98 \), \( a^3 = 4 \), and \( a^2 = -10 \):

\[
-98 = -4(4) + 5(-10) - 4a^1
\]

Simplify the terms:

\[
-98 = -16 - 50 - 4a^1
\]

\[
-98 = -66 - 4a^1
\]

\[
-98 + 66 = -4a^1
\]

\[
-32 = -4a^1
\]

\[
a^1 = 8
\]

### Final Answer

Thus, the values of \( a^1 \), \( a^2 \), and \( a^3 \) are:

\[
a^1 = 8, \quad a^2 = -10, \quad a^3 = 4
\]

Would you like a more detailed explanation or have any questions?

---

Here are 5 related questions:

1. How can you generalize recurrence relations for higher-order terms?
2. What are other methods to solve recurrence relations besides substitution?
3. How would solving change if the relation included a constant term?
4. How can recurrence relations model real-world problems like population growth?
5. What role do initial conditions play in determining solutions to recurrence relations?

**Tip:** In recurrence relations, always pay attention to the order of terms—it directly impacts how the sequence evolves.

Learn how to solve the recurrence relation a^n = -4a^(n-1) + 5a^(n-2) - 4a^(n-3) with given values a^4 = -98, a^5 = 452, and a^6 = -2314. This step-by-step guide explains how to find a^1, a^2, and a^3. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation