recurrence relation  𝑎 𝑛 − 2 𝑎 𝑛 − 1 + 2 𝑎 𝑛 − 2 = 5 a  n ​  −2a  n−1 ​  +2a  n−2 ​  =5
Solve the question again but remove the real valued trigonometric form

Given the recurrence relation:

\[
a_n - 2a_{n-1} + 2a_{n-2} = 5
\]

We start by solving the corresponding homogeneous equation:

\[
a_n - 2a_{n-1} + 2a_{n-2} = 0
\]

### Step 1: Solve the Characteristic Equation

The characteristic equation associated with this recurrence relation is:

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

This is a quadratic equation, so we solve for \( r \) using the quadratic formula:

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

Here, \( a = 1 \), \( b = -2 \), and \( c = 2 \). Plugging these values into the formula:

\[
r = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2}
\]

Thus, the roots are:

\[
r_1 = 1 + i, \quad r_2 = 1 - i
\]

### Step 2: General Solution to the Homogeneous Equation

The general solution to the homogeneous equation is:

\[
a_n^{(h)} = c_1(1 + i)^n + c_2(1 - i)^n
\]

### Step 3: Find the Particular Solution

Given that the non-homogeneous term is a constant (5), we look for a particular solution \( a_n^{(p)} \) that is also constant. Let:

\[
a_n^{(p)} = A
\]

Substituting this into the original recurrence relation:

\[
A - 2A + 2A = 5
\]

Simplifying, we get:

\[
A = 5
\]

So, the particular solution is:

\[
a_n^{(p)} = 5
\]

### Step 4: General Solution to the Non-Homogeneous Recurrence Relation

The general solution to the non-homogeneous recurrence relation is the sum of the homogeneous and particular solutions:

\[
a_n = a_n^{(h)} + a_n^{(p)} = c_1(1 + i)^n + c_2(1 - i)^n + 5
\]

### Step 5: Simplify the Complex Roots

Now, instead of using the trigonometric form, we express \( (1 \pm i)^n \) directly:

For \( z = 1 + i \), the magnitude \( |z| = \sqrt{2} \) and argument \( \theta = \frac{\pi}{4} \), but we avoid trigonometric terms by directly simplifying:

\[
(1 + i)^n = (\sqrt{2})^n \left(\cos\left(\frac{n\pi}{4}\right) + i\sin\left(\frac{n\pi}{4}\right)\right)
\]

Expanding without trigonometric form, this translates to the sum of real and imaginary parts, which is not straightforwardly simplified without reverting to trigonometric or exponential forms. However, since the user requests avoiding trigonometric forms, we leave it in its expanded complex form:

\[
a_n = c_1(1 + i)^n + c_2(1 - i)^n + 5
\]

Where \( (1 + i)^n \) and \( (1 - i)^n \) can be expanded using binomial theorem if needed.

This is the solution without expressing the complex roots in their trigonometric form.

---

Would you like to explore this further or have any questions? Here are some related questions you might consider:

1. How would the solution change if the non-homogeneous term was a function of \( n \)?
2. Can you express the real part of the solution explicitly without using trigonometric identities?
3. How does the particular solution form change if the right side is \( 5n \) instead of \( 5 \)?
4. What if the characteristic roots were real and distinct? How would the solution differ?
5. How does the initial condition influence the constants \( c_1 \) and \( c_2 \) in the final solution?

**Tip:** Complex roots in recurrence relations often suggest oscillatory behavior, which can be expressed using sine and cosine terms, but they can also be expanded using binomials for non-oscillatory interpretations.

Learn how to solve the recurrence relation a_n - 2a_{n-1} + 2a_{n-2} = 5 with a detailed step-by-step solution. This problem involves complex numbers, homogeneous equations, and the binomial theorem. Suitable for advanced high school students and beyond.

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