To prove $a_n = 3 \cdot 2^n - 2 \cdot 3^n$ is a closed form of the recurrence relation $a_n = 5a_{n-1} - 6a_{n-2}$ with initial conditions $a_0 = 1$ and $a_1 = 0$, we use strong mathematical induction.

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Steps:

Base Cases

1. For $n = 0$: Substitute $n = 0$ into the closed form: $a_0 = 3 \cdot 2^0 - 2 \cdot 3^0 = 3 \cdot 1 - 2 \cdot 1 = 1$ This matches $a_0 = 1$.

2. For $n = 1$: Substitute $n = 1$ into the closed form: $a_1 = 3 \cdot 2^1 - 2 \cdot 3^1 = 3 \cdot 2 - 2 \cdot 3 = 6 - 6 = 0$ This matches $a_1 = 0$.

Thus, the base cases hold.

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Inductive Hypothesis

Assume the closed form $a_k = 3 \cdot 2^k - 2 \cdot 3^k$ is true for $k = n$ and $k = n-1$, i.e., $a_n = 3 \cdot 2^n - 2 \cdot 3^n$ $a_{n-1} = 3 \cdot 2^{n-1} - 2 \cdot 3^{n-1}$

We aim to prove that $a_{n+1} = 3 \cdot 2^{n+1} - 2 \cdot 3^{n+1}$ satisfies the recurrence relation.

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Inductive Step

From the recurrence relation $a_{n+1} = 5a_n - 6a_{n-1}$, substitute the inductive hypothesis into the right-hand side:

$a_{n+1} = 5(3 \cdot 2^n - 2 \cdot 3^n) - 6(3 \cdot 2^{n-1} - 2 \cdot 3^{n-1})$

Simplify each term:

1. Expand the first term: $5(3 \cdot 2^n - 2 \cdot 3^n) = 15 \cdot 2^n - 10 \cdot 3^n$

2. Expand the second term: $-6(3 \cdot 2^{n-1} - 2 \cdot 3^{n-1}) = -18 \cdot 2^{n-1} + 12 \cdot 3^{n-1}$

   Recall that $2^{n-1} = \frac{2^n}{2}$ and $3^{n-1} = \frac{3^n}{3}$: $-18 \cdot 2^{n-1} = -9 \cdot 2^n, \quad 12 \cdot 3^{n-1} = 4 \cdot 3^n$

Thus: $a_{n+1} = (15 \cdot 2^n - 10 \cdot 3^n) + (-9 \cdot 2^n + 4 \cdot 3^n)$

Combine like terms: $a_{n+1} = (15 \cdot 2^n - 9 \cdot 2^n) + (-10 \cdot 3^n + 4 \cdot 3^n)$ $a_{n+1} = 6 \cdot 2^n - 6 \cdot 3^n$

Factor out $6$: $a_{n+1} = 3 \cdot 2^{n+1} - 2 \cdot 3^{n+1}$

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Conclusion

We have shown that $a_{n+1} = 3 \cdot 2^{n+1} - 2 \cdot 3^{n+1}$, completing the inductive step. By the principle of strong mathematical induction, $a_n = 3 \cdot 2^n - 2 \cdot 3^n$ is the closed form solution for all $n \geq 0$.

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Would you like me to elaborate on any step?

Expanding Ideas:

1. What are common methods for solving recurrence relations?
2. How can characteristic equations help in deriving closed forms?
3. Can generating functions solve the same recurrence relation?
4. What are the advantages of using induction for proving solutions?
5. How does this result apply to combinatorial problems?

Tip:

Always verify the base cases and write down all assumptions explicitly when using induction.