Answer
$a_k
= (\sqrt{2})^k\,\Bigl[\cos\!\bigl(k\tfrac{\pi}{4}\bigr)
+ \sin\!\bigl(k\tfrac{\pi}{4}\bigr)\Bigr],
\quad k\ge0.
$
Work Step by Step
**Solution Explanation**
We want to solve the second‐order linear recurrence
\[
a_k \;=\; 2\,a_{k-1} \;-\; 2\,a_{k-2},
\quad \text{for }k\ge2,
\quad \text{with }a_0 = 1,\; a_1 = 2.
\]
---
## 1. Characteristic Equation
We look for solutions of the form \(a_k = r^k\). Substituting into the recurrence:
\[
r^k
\;=\; 2\,r^{k-1} \;-\; 2\,r^{k-2}.
\]
Divide both sides by \(r^{k-2}\) (assuming \(r\neq 0\)):
\[
r^2
= 2\,r \;-\; 2,
\]
which rearranges to
\[
r^2 \;-\; 2\,r \;+\; 2 \;=\; 0.
\]
The discriminant is \(\Delta = (-2)^2 - 4(1)(2) = 4 - 8 = -4\), which is **negative**. Hence the roots are complex conjugates:
\[
r \;=\; \frac{2 \pm \sqrt{-4}}{2}
\;=\; \frac{2 \pm 2i}{2}
\;=\; 1 \;\pm\; i.
\]
So the two roots are \(r_1 = 1 + i\) and \(r_2 = 1 - i\).
---
## 2. Converting to a Real‐Valued Solution
Because the recurrence and initial conditions are real, the actual solution \(\{a_k\}\) must be real for all \(k\). We know from theory that for complex‐conjugate roots \(p \pm qi\), one standard real form is
\[
a_k = R^k \,\Bigl[\alpha\,\cos(k\theta) + \beta\,\sin(k\theta)\Bigr],
\]
where \(R = \sqrt{p^2 + q^2}\) and \(\theta = \tan^{-1}\!\bigl(q/p\bigr)\).
In our case, \(p=1\), \(q=1\). So
1. \(R = \sqrt{1^2 + 1^2} = \sqrt{2}.\)
2. \(\theta = \tan^{-1}(1/1) = \pi/4.\)
Hence the general real solution can be written:
\[
a_k
= (\sqrt{2})^{\,k}\,\bigl[\alpha\,\cos\!\bigl(k\tfrac{\pi}{4}\bigr)
\;+\; \beta\,\sin\!\bigl(k\tfrac{\pi}{4}\bigr)\bigr].
\]
We will determine \(\alpha\) and \(\beta\) from the initial conditions \(a_0=1\) and \(a_1=2\).
---
## 3. Use Initial Conditions to Solve for \(\alpha\) and \(\beta\)
### Condition at \(k=0\)
\[
a_0 = (\sqrt{2})^0
\bigl[\alpha\,\cos(0) + \beta\,\sin(0)\bigr]
= \alpha \cdot 1 + \beta \cdot 0
= \alpha.
\]
But \(a_0=1\). Hence
\[
\alpha = 1.
\]
### Condition at \(k=1\)
\[
a_1
= (\sqrt{2})^1
\bigl[\alpha\,\cos\!\bigl(\tfrac{\pi}{4}\bigr)
+ \beta\,\sin\!\bigl(\tfrac{\pi}{4}\bigr)\bigr].
\]
We know \(a_1=2\), \(\alpha=1\), and \(\cos(\pi/4) = \sin(\pi/4)=\tfrac{\sqrt{2}}{2}\). So:
\[
2
= \sqrt{2}\,\bigl[\,1\cdot\tfrac{\sqrt{2}}{2}
+ \beta\cdot\tfrac{\sqrt{2}}{2}\bigr]
= \sqrt{2}\,\cdot \tfrac{\sqrt{2}}{2}\,\bigl[\,1 + \beta\,\bigr]
= 1 \cdot (1 + \beta).
\]
Hence \(1 + \beta = 2\), so \(\beta=1\).
---
## 4. Final Closed‐Form Solution
Putting \(\alpha=1\) and \(\beta=1\) back in, we get
\[
\boxed{
a_k
= (\sqrt{2})^k\,\Bigl[\cos\!\bigl(k\tfrac{\pi}{4}\bigr)
+ \sin\!\bigl(k\tfrac{\pi}{4}\bigr)\Bigr],
\quad k\ge0.
}
\]
That is a perfectly good real‐form solution. One can also express \(\cos x + \sin x\) in other trigonometric identities if desired, but the above form is standard and satisfies:
\(a_0=1\),
\(a_1=2\),
\(a_k\in\mathbb{R}\) for all \(k\),
and the given recurrence.
---
### Optional Check of the First Few Terms
- \(k=0\): \(a_0 = 2^0 \bigl[\cos(0)+\sin(0)\bigr] = 1+0=1.\)
- \(k=1\): \(a_1 = \sqrt{2}\bigl[\cos(\pi/4)+\sin(\pi/4)\bigr]
= \sqrt{2}\bigl[\tfrac{\sqrt{2}}{2} + \tfrac{\sqrt{2}}{2}\bigr]
= \sqrt{2}\cdot \sqrt{2} = 2.\)
Hence it matches the initial conditions as required.
