Answer
\[
\boxed{s_k \;=\; \frac{k}{2}\,(-2)^k, \quad \text{for all integers }k\ge0.}
\]
Work Step by Step
We want to solve the recurrence
\[
s_k \;=\; -4\,s_{k-1} \;-\; 4\,s_{k-2},
\quad \text{for }k\ge2,
\quad\text{with } s_0 = 0,\; s_1 = -1.
\]
---
## 1. Characteristic Equation
Look for solutions of the form \(s_k = t^k\). Substituting into the recurrence:
\[
t^k
\;=\; -4\,t^{k-1} \;-\; 4\,t^{k-2}.
\]
Divide both sides by \(t^{k-2}\) (assuming \(t \neq 0\)):
\[
t^2
= -4\,t \;-\; 4,
\]
which rearranges to
\[
t^2 + 4\,t + 4 \;=\; 0.
\]
Factor or use the quadratic formula. Notice it is a perfect square:
\[
(t + 2)^2 \;=\; 0
\quad\Longrightarrow\quad
t = -2 \quad\text{(a repeated root).}
\]
---
## 2. General Solution for a Repeated Root
When the characteristic equation has a double root \(r\), the general homogeneous solution takes the form
\[
s_k
= A \,r^k + B\,k\,r^k.
\]
Here, \(r = -2\). Hence
\[
s_k
= A\,(-2)^k + B\,k\,(-2)^k.
\]
---
## 3. Determine \(A\) and \(B\) from Initial Conditions
We have:
1. \(s_0 = 0\).
2. \(s_1 = -1\).
Substitute:
- **For \(k=0\)**:
\[
s_0
= A\,(-2)^0 + B\,(0)\,(-2)^0
= A \;=\; 0.
\]
So \(A=0\).
- **For \(k=1\)**:
\[
s_1
= A\,(-2)^1 + B\,(1)\,(-2)^1
= A\cdot(-2) + B\cdot(-2).
\]
But \(A=0\), so this reduces to
\[
s_1 = -2\,B = -1
\quad\Longrightarrow\quad
B = \tfrac{1}{2}.
\]
Thus \(A=0\) and \(B=\tfrac{1}{2}\).
---
## 4. Final Closed‐Form Solution
Plugging back into \(s_k = A\,(-2)^k + B\,k\,(-2)^k\), with \(A=0\) and \(B=\tfrac{1}{2}\), we get
\[
s_k
= \frac{1}{2}\,k\,(-2)^k.
\]
