Solve the RR an=20an-1 -25an-2 for n>=2 where a0=0 and a1=1. Solve correctly and properly

To solve the recurrence relation \( a_n = 20a_{n-1} - 25a_{n-2} \) for \( n \geq 2 \) with initial conditions \( a_0 = 0 \) and \( a_1 = 1 \), we can follow these steps:

### Step 1: Write down the characteristic equation
The recurrence relation is a linear homogeneous recurrence relation with constant coefficients. The corresponding characteristic equation is:

\[
r^2 = 20r - 25
\]

### Step 2: Solve the characteristic equation
Rearrange the characteristic equation to get:

\[
r^2 - 20r + 25 = 0
\]

This is a quadratic equation, which we can solve using the quadratic formula:

\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -20 \), and \( c = 25 \). Plugging in these values:

\[
r = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(25)}}{2(1)}
\]

\[
r = \frac{20 \pm \sqrt{400 - 100}}{2}
\]

\[
r = \frac{20 \pm \sqrt{300}}{2}
\]

\[
r = \frac{20 \pm 10\sqrt{3}}{2}
\]

\[
r = 10 \pm 5\sqrt{3}
\]

So, the roots of the characteristic equation are \( r_1 = 10 + 5\sqrt{3} \) and \( r_2 = 10 - 5\sqrt{3} \).

### Step 3: Write the general solution
The general solution of the recurrence relation is given by:

\[
a_n = A \cdot r_1^n + B \cdot r_2^n
\]

Substituting \( r_1 = 10 + 5\sqrt{3} \) and \( r_2 = 10 - 5\sqrt{3} \), we have:

\[
a_n = A \cdot (10 + 5\sqrt{3})^n + B \cdot (10 - 5\sqrt{3})^n
\]

### Step 4: Use initial conditions to find constants \( A \) and \( B \)
We use the initial conditions \( a_0 = 0 \) and \( a_1 = 1 \) to solve for \( A \) and \( B \).

**For \( a_0 = 0 \):**

\[
a_0 = A \cdot (10 + 5\sqrt{3})^0 + B \cdot (10 - 5\sqrt{3})^0
\]

\[
0 = A + B
\]

So, \( B = -A \).

**For \( a_1 = 1 \):**

\[
a_1 = A \cdot (10 + 5\sqrt{3}) + B \cdot (10 - 5\sqrt{3})
\]

Substitute \( B = -A \):

\[
1 = A \cdot (10 + 5\sqrt{3}) - A \cdot (10 - 5\sqrt{3})
\]

Factor out \( A \):

\[
1 = A \cdot \left[(10 + 5\sqrt{3}) - (10 - 5\sqrt{3})\right]
\]

\[
1 = A \cdot \left[10 + 5\sqrt{3} - 10 + 5\sqrt{3}\right]
\]

\[
1 = A \cdot \left[10\sqrt{3}\right]
\]

\[
A = \frac{1}{10\sqrt{3}}
\]

Thus, \( B = -\frac{1}{10\sqrt{3}} \).

### Step 5: Write the final solution
Substituting the values of \( A \) and \( B \):

\[
a_n = \frac{1}{10\sqrt{3}} \cdot (10 + 5\sqrt{3})^n - \frac{1}{10\sqrt{3}} \cdot (10 - 5\sqrt{3})^n
\]

Simplify the expression:

\[
a_n = \frac{(10 + 5\sqrt{3})^n - (10 - 5\sqrt{3})^n}{10\sqrt{3}}
\]

This is the closed-form solution for the recurrence relation.

Would you like a more detailed explanation of any part of this solution? Here are some related questions to consider:

1. What are the general steps for solving linear homogeneous recurrence relations?
2. How does the characteristic equation relate to the recurrence relation?
3. Can you derive the quadratic formula used in solving the characteristic equation?
4. What is the significance of the initial conditions in determining the solution?
5. How does the nature of the roots (real, complex, repeated) affect the general solution?
6. How do you verify that the solution satisfies the original recurrence relation?
7. What would the solution look like if the initial conditions were different?
8. How would the method change if the recurrence relation had non-homogeneous terms?

**Tip:** Always verify your final solution by substituting it back into the original recurrence relation. This ensures that your derived formula is correct.

Learn how to solve the recurrence relation a_n = 20a_{n-1} - 25a_{n-2} with initial conditions a_0 = 0 and a_1 = 1. Follow a detailed step-by-step solution using the characteristic equation method and quadratic formula. Suitable for advanced high school students.

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