To solve the recurrence relation

$a_n = 7a_{n-1} - 10a_{n-2}$

with initial conditions $a_0 = 2$ and $a_1 = 7$, we can approach it using the characteristic equation method.

Step 1: Find the Characteristic Equation

The characteristic equation for a recurrence relation of the form $a_n = c_1 a_{n-1} + c_2 a_{n-2}$ is:

$x^2 - c_1 x - c_2 = 0$

For our recurrence, this becomes:

$x^2 - 7x + 10 = 0$

Step 2: Solve the Characteristic Equation

To find the roots of the equation $x^2 - 7x + 10 = 0$, we use the quadratic formula:

$x = \frac{7 \pm \sqrt{7^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm \sqrt{9}}{2}$

This gives:

$x = \frac{7 \pm 3}{2}$

The roots are:

$x = 5 \quad \text{and} \quad x = 2$

Step 3: Write the General Solution

Since we have distinct roots, the general solution for $a_n$ is:

$a_n = A \cdot 5^n + B \cdot 2^n$

where $A$ and $B$ are constants determined by the initial conditions.

Step 4: Apply Initial Conditions

1. Using $a_0 = 2$:

   $a_0 = A \cdot 5^0 + B \cdot 2^0 = A + B = 2$

2. Using $a_1 = 7$:

   $a_1 = A \cdot 5^1 + B \cdot 2^1 = 5A + 2B = 7$

Now we have the system of equations:

$A + B = 2$ $5A + 2B = 7$

Step 5: Solve for $A$ and $B$

From the first equation, $B = 2 - A$.

Substitute $B$ into the second equation:

$5A + 2(2 - A) = 7$ $5A + 4 - 2A = 7$ $3A = 3$ $A = 1$

Now substitute $A = 1$ back into $B = 2 - A$:

$B = 2 - 1 = 1$

Final Solution

The solution to the recurrence relation is:

$a_n = 1 \cdot 5^n + 1 \cdot 2^n = 5^n + 2^n$

Thus,

$a_n = 5^n + 2^n$

Would you like additional details or have any questions about this solution?

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Here are some related questions to further explore this topic:

1. What if we change the initial conditions? How would it affect the solution?
2. How do we solve recurrence relations with complex roots?
3. Can this approach be applied to higher-order recurrence relations?
4. How does the solution change for a non-homogeneous recurrence relation?
5. What are alternative methods to solve recurrence relations besides the characteristic equation method?

Tip: When dealing with recurrence relations, always verify the solution by plugging back the initial conditions to ensure accuracy.