Find the solution to the recurrence relation an=-an-1+12an-2 with initial terms a0=2 and a1=-1.                                  an=

To solve the recurrence relation \( a_n = -a_{n-1} + 12a_{n-2} \) with initial terms \( a_0 = 2 \) and \( a_1 = -1 \), we will follow these steps:

### Step 1: Write the characteristic equation
This is a linear recurrence relation with constant coefficients, so we can assume the solution takes the form \( a_n = r^n \), where \( r \) is a constant to be determined. Substituting \( a_n = r^n \) into the recurrence relation:

\[
r^n = -r^{n-1} + 12r^{n-2}
\]

Dividing both sides by \( r^{n-2} \) (assuming \( r \neq 0 \)):

\[
r^2 = -r + 12
\]

Rearranging:

\[
r^2 + r - 12 = 0
\]

### Step 2: Solve the characteristic equation
We now solve the quadratic equation \( r^2 + r - 12 = 0 \). We can use the quadratic formula:

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

Here, \( a = 1 \), \( b = 1 \), and \( c = -12 \):

\[
r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)}
\]
\[
r = \frac{-1 \pm \sqrt{1 + 48}}{2}
\]
\[
r = \frac{-1 \pm \sqrt{49}}{2}
\]
\[
r = \frac{-1 \pm 7}{2}
\]

Thus, the two roots are:

\[
r_1 = \frac{-1 + 7}{2} = 3 \quad \text{and} \quad r_2 = \frac{-1 - 7}{2} = -4
\]

### Step 3: General solution
The general solution to the recurrence relation is:

\[
a_n = A \cdot 3^n + B \cdot (-4)^n
\]

where \( A \) and \( B \) are constants to be determined by the initial conditions.

### Step 4: Apply the initial conditions
We use the initial conditions \( a_0 = 2 \) and \( a_1 = -1 \) to find \( A \) and \( B \).

For \( a_0 = 2 \):

\[
2 = A \cdot 3^0 + B \cdot (-4)^0
\]
\[
2 = A + B
\]

For \( a_1 = -1 \):

\[
-1 = A \cdot 3^1 + B \cdot (-4)^1
\]
\[
-1 = 3A - 4B
\]

Now we solve this system of equations:

1. \( A + B = 2 \)
2. \( 3A - 4B = -1 \)

### Step 5: Solve the system of equations
From the first equation, \( A = 2 - B \), substitute this into the second equation:

\[
3(2 - B) - 4B = -1
\]
\[
6 - 3B - 4B = -1
\]
\[
6 - 7B = -1
\]
\[
-7B = -7
\]
\[
B = 1
\]

Substitute \( B = 1 \) into \( A + B = 2 \):

\[
A + 1 = 2
\]
\[
A = 1
\]

### Step 6: Final solution
Thus, the solution to the recurrence relation is:

\[
a_n = 1 \cdot 3^n + 1 \cdot (-4)^n
\]
\[
a_n = 3^n + (-4)^n
\]

### Conclusion
The general solution to the recurrence relation is:

\[
a_n = 3^n + (-4)^n
\]

---

Would you like any further clarification on the solution, or have any questions about the process? Here are 5 related questions for further exploration:

1. How would the solution change if the recurrence relation had different coefficients, like \( 2a_{n-1} + 5a_{n-2} \)?
2. How can we solve recurrence relations that do not have constant coefficients?
3. What is the significance of the characteristic equation in solving linear recurrence relations?
4. Can recurrence relations be solved using generating functions? How would that method work for this problem?
5. What would happen if the recurrence relation had complex roots instead of real ones?

**Tip:** When solving recurrence relations, always start by determining the characteristic equation, as it gives insight into the general form of the solution, especially for constant-coefficient linear recurrences.

Find the solution to the recurrence relation an=-an-1+12an-2 with initial terms a0=2 and a1=-1.

Learn how to solve the recurrence relation an = -an-1 + 12an-2 with initial conditions a0 = 2 and a1 = -1. This step-by-step solution demonstrates the use of the characteristic equation, quadratic formula, and general solution for linear recurrences. Suitable for high school students in Grades 10-12.

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