The recurrence relation provided is:

$a_n = a_{n-1} + 2a_{n-2}, \quad \text{with } a_0 = 2 \text{ and } a_1 = 7.$

To solve this recurrence relation:

1. Find the characteristic equation:
   The homogeneous recurrence relation is $a_n = a_{n-1} + 2a_{n-2}$, which can be transformed into the characteristic equation: $r^n = r^{n-1} + 2r^{n-2}$ Dividing both sides by $r^{n-2}$, we get: $r^2 = r + 2$ Rearranging terms: $r^2 - r - 2 = 0$ This quadratic equation can be factored as: $(r - 2)(r + 1) = 0$ Hence, the solutions to the characteristic equation are $r_1 = 2$ and $r_2 = -1$.

2. General solution:
   The general solution to the recurrence relation is of the form: $a_n = c_1 \cdot 2^n + c_2 \cdot (-1)^n$ where $c_1$ and $c_2$ are constants to be determined from the initial conditions.

3. Use the initial conditions to solve for $c_1$ and $c_2$:

   • From $a_0 = 2$: $a_0 = c_1 \cdot 2^0 + c_2 \cdot (-1)^0 = c_1 + c_2 = 2$
   • From $a_1 = 7$: $a_1 = c_1 \cdot 2^1 + c_2 \cdot (-1)^1 = 2c_1 - c_2 = 7$

4. Solve the system of equations:

   c_1 + c_2 = 2 \\ 2c_1 - c_2 = 7 \end{cases}$$ Add both equations: $$3c_1 = 9 \quad \Rightarrow \quad c_1 = 3$$ Substitute $$c_1 = 3$$ into the first equation: $$3 + c_2 = 2 \quad \Rightarrow \quad c_2 = -1$$

5. Final solution: The solution to the recurrence relation is: $a_n = 3 \cdot 2^n - (-1)^n$

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Related questions:

1. What are characteristic equations, and how do they help in solving recurrence relations?
2. Can this method be applied to non-homogeneous recurrence relations?
3. How does changing the initial conditions affect the solution of a recurrence relation?
4. What is the difference between linear and non-linear recurrence relations?
5. How do you solve recurrence relations with repeated roots in their characteristic equations?

Tip:

When solving recurrence relations, always verify that your characteristic equation has real roots before proceeding with the general solution.