Answer
$b_k
= (\sqrt{5})^{\,k}\,\cos\!\Bigl(k \,\arctan(2)\Bigr),
\quad k \ge 0$
Work Step by Step
We want to solve the second‐order linear recurrence
\[
b_k \;=\; 2\,b_{k-1} \;-\; 5\,b_{k-2},
\quad k \ge 2,
\quad \text{with}\quad b_0 = 1,\; b_1 = 1.
\]
---
## 1. Characteristic Equation
We look for solutions of the form \(b_k = r^k\). Substituting into the recurrence:
\[
r^k
\;=\; 2\,r^{k-1} \;-\; 5\,r^{k-2}.
\]
Divide both sides by \(r^{k-2}\) (assuming \(r\neq 0\)):
\[
r^2
= 2\,r \;-\; 5,
\]
which rearranges to
\[
r^2 \;-\; 2\,r \;+\; 5 \;=\; 0.
\]
The discriminant is \(\Delta = (-2)^2 - 4(1)(5) = 4 - 20 = -16 < 0\).
Thus, the roots are **complex conjugates**:
\[
r
= \frac{2 \pm \sqrt{-16}}{2}
= \frac{2 \pm 4i}{2}
= 1 \;\pm\; 2i.
\]
---
## 2. Converting to a Real‐Valued Form
Since the recurrence and initial conditions are real, the actual sequence \(\{b_k\}\) must be real. For roots of the form \(p \pm q\,i\), a standard real solution is
\[
b_k
= R^k \Bigl[\alpha \,\cos(k\theta) \;+\; \beta \,\sin(k\theta)\Bigr],
\]
where \(R = \sqrt{p^2 + q^2}\) and \(\theta = \arctan\!\bigl(\tfrac{q}{p}\bigr)\).
Here, \(p = 1\) and \(q = 2\). So:
1. \(R = \sqrt{1^2 + 2^2} = \sqrt{5}.\)
2. \(\theta = \arctan\!\bigl(\tfrac{2}{1}\bigr) = \arctan(2).\)
Hence the general real solution can be written:
\[
b_k
= (\sqrt{5})^k \Bigl[\alpha \,\cos\bigl(k\,\theta\bigr)
+ \beta \,\sin\bigl(k\,\theta\bigr)\Bigr],
\quad
\text{where}\quad \theta = \arctan(2).
\]
We will determine \(\alpha\) and \(\beta\) using the initial conditions \(b_0=1\) and \(b_1=1\).
---
## 3. Use the Initial Conditions to Solve for \(\alpha\) and \(\beta\)
### (a) At \(k=0\)
\[
b_0
= (\sqrt{5})^0 \Bigl[\alpha \,\cos(0) + \beta \,\sin(0)\Bigr]
= \alpha.
\]
Since \(b_0=1\), we get
\[
\alpha = 1.
\]
### (b) At \(k=1\)
\[
b_1
= (\sqrt{5})^1 \Bigl[\alpha\,\cos(\theta) + \beta\,\sin(\theta)\Bigr].
\]
We know \(b_1=1\), and we have \(\alpha=1\). Also, \(\cos(\theta) = \cos(\arctan(2))\) and \(\sin(\theta)=\sin(\arctan(2))\). Recall that if \(\theta=\arctan(2)\), then
\[
\cos(\theta) = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}},
\quad
\sin(\theta) = \frac{2}{\sqrt{5}}.
\]
Hence
\[
1
= \sqrt{5}\,\Bigl[1 \cdot \tfrac{1}{\sqrt{5}}
+ \beta \cdot \tfrac{2}{\sqrt{5}}\Bigr]
= \sqrt{5}\,\cdot \tfrac{1}{\sqrt{5}}\,
\Bigl[\,1 + 2\beta\Bigr]
= 1 \cdot \bigl(1 + 2\beta\bigr).
\]
Therefore \(1 + 2\beta = 1\), implying \(\beta = 0\).
---
## 4. Final Closed‐Form Solution
Putting \(\alpha=1\) and \(\beta=0\) back into the general real solution:
\[
b_k
= (\sqrt{5})^k \Bigl[\cos\bigl(k\,\arctan(2)\bigr)\Bigr].
\]
Hence,
\[
\boxed{
b_k
= (\sqrt{5})^{\,k}\,\cos\!\Bigl(k \,\arctan(2)\Bigr),
\quad k \ge 0.
}
\]
This satisfies both the recurrence and the initial conditions \(b_0=1\), \(b_1=1\).
