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?

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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.