Answer
\[
\boxed{
s_k
= \frac{3 + 2\sqrt{3}}{6}\,\bigl(1 + \sqrt{3}\bigr)^k
\;+\;
\frac{3 - 2\sqrt{3}}{6}\,\bigl(1 - \sqrt{3}\bigr)^k,
}
\quad k \ge 0.
\]
Work Step by Step
We have the second‐order linear recurrence
\[
s_k \;=\; 2\,s_{k-1} \;+\; 2\,s_{k-2},
\quad \text{for }k \ge 2,
\quad \text{with } s_0 = 1,\; s_1 = 3.
\]
Below is a step‐by‐step solution to find its closed‐form formula.
---
## 1. Characteristic Equation
We look for solutions of the form \(s_k = r^k\). Substituting into the recurrence:
\[
r^k
\;=\; 2\,r^{k-1} \;+\; 2\,r^{k-2}.
\]
Divide by \(r^{k-2}\) (assuming \(r\neq0\)):
\[
r^2
= 2\,r \;+\; 2,
\]
which rearranges to
\[
r^2 \;-\; 2\,r \;-\; 2 = 0.
\]
This is the **characteristic equation**. Let’s solve it using the quadratic formula:
\[
r
= \frac{2 \pm \sqrt{(2)^2 \;+\; 4 \cdot 2}}{2}
= \frac{2 \pm \sqrt{4 + 8}}{2}
= \frac{2 \pm \sqrt{12}}{2}
= \frac{2 \pm 2\sqrt{3}}{2}
= 1 \;\pm\; \sqrt{3}.
\]
Hence the two (distinct) roots are
\[
r_1 = 1 + \sqrt{3},
\quad
r_2 = 1 - \sqrt{3}.
\]
---
## 2. General Form of the Solution
Because the characteristic equation has two distinct real roots \(r_1\) and \(r_2\), the general solution of the recurrence is
\[
s_k = A\,(1 + \sqrt{3})^k \;+\; B\,(1 - \sqrt{3})^k,
\]
for some constants \(A\) and \(B\).
---
## 3. Determine \(A\) and \(B\) Using the Initial Conditions
We have:
\[
s_0 = 1,
\quad
s_1 = 3.
\]
### Condition at \(k=0\)
\[
s_0
= A\,(1 + \sqrt{3})^0 + B\,(1 - \sqrt{3})^0
= A + B
= 1.
\]
So
\[
(1)\quad A + B = 1.
\]
### Condition at \(k=1\)
\[
s_1
= A\,(1 + \sqrt{3})^1 + B\,(1 - \sqrt{3})^1
= A\,(1 + \sqrt{3}) + B\,(1 - \sqrt{3})
= 3.
\]
So
\[
(2)\quad A(1 + \sqrt{3}) + B(1 - \sqrt{3}) = 3.
\]
---
## 4. Solve for \(A\) and \(B\)
From equation (1), \(B = 1 - A\). Substitute into (2):
\[
A(1 + \sqrt{3})
\;+\; \bigl(1 - A\bigr)\,(1 - \sqrt{3})
= 3.
\]
Expand:
\[
A(1 + \sqrt{3})
\;+\; \bigl[\,1 - \sqrt{3}\bigr]
\;-\; A(1 - \sqrt{3})
= 3.
\]
Group the \(A\)‐terms carefully:
- \(A(1 + \sqrt{3}) - A(1 - \sqrt{3})
= A\bigl[(1 + \sqrt{3}) - (1 - \sqrt{3})\bigr]
= A\bigl(\sqrt{3} + \sqrt{3}\bigr)
= 2\sqrt{3}\,A.\)
Hence the entire left side becomes
\[
2\sqrt{3}\,A
\;+\; (1 - \sqrt{3}).
\]
So the equation is
\[
2\sqrt{3}\,A + (1 - \sqrt{3}) = 3.
\]
Isolate \(A\):
\[
2\sqrt{3}\,A
= 3 - (1 - \sqrt{3})
= 3 - 1 + \sqrt{3}
= 2 + \sqrt{3}.
\]
Therefore
\[
A
= \frac{2 + \sqrt{3}}{2\sqrt{3}}.
\]
You can leave \(A\) in this form or simplify by multiplying numerator and denominator by \(\sqrt{3}\):
\[
A
= \frac{2 + \sqrt{3}}{2\sqrt{3}}
\,\times\,
\frac{\sqrt{3}}{\sqrt{3}}
= \frac{(2 + \sqrt{3})\,\sqrt{3}}{2\cdot 3}
= \frac{2\sqrt{3} + 3}{6}.
\]
Either form is correct. Let’s write
\[
A = \frac{3 + 2\sqrt{3}}{6}.
\]
Since \(B = 1 - A\),
\[
B
= 1 - \frac{3 + 2\sqrt{3}}{6}
= \frac{6 - (3 + 2\sqrt{3})}{6}
= \frac{3 - 2\sqrt{3}}{6}.
\]
able as long as it clearly shows the dependence on \(k\).
