Answer
\[
\boxed{
e_k
= \begin{cases}
0, & \text{if $k$ is even},\\[6pt]
2 \cdot 9^{\tfrac{k-1}{2}}, & \text{if $k$ is odd}.
\end{cases}
}
\]
Work Step by Step
We have the recurrence
\[
e_k \;=\; 9\,e_{k-2},
\quad \text{for all integers }k \ge 2,
\quad\text{with}\quad e_0=0,\;e_1=2.
\]
Because each term depends only on the term two steps before it, we can separate the sequence into its even‐indexed and odd‐indexed parts.
---
## 1. Compute the first few terms
\(e_0 = 0\) (given)
\(e_1 = 2\) (given)
\(e_2 = 9\,e_0 = 9\cdot 0 = 0\)
\(e_3 = 9\,e_1 = 9\cdot 2 = 18\)
\(e_4 = 9\,e_2 = 9\cdot 0 = 0\)
\(e_5 = 9\,e_3 = 9\cdot 18 = 162\)
\(e_6 = 9\,e_4 = 9\cdot 0 = 0\)
\(e_7 = 9\,e_5 = 9\cdot 162 = 1458\)
\(\dots\)
From these values:
- **Even indices**: \(e_0 = 0,\; e_2=0,\; e_4=0,\dots\) all zero.
- **Odd indices**: \(e_1=2,\; e_3=18,\; e_5=162,\; e_7=1458,\dots\)
Notice that for odd \(k = 2n+1\):
\[
e_{2n+1} = 2,\;18,\;162,\;1458,\dots
\]
We see each is multiplied by 9 as \(n\) increases by 1. Indeed,
\[
e_1 = 2 \times 9^0,\quad
e_3 = 2 \times 9^1,\quad
e_5 = 2 \times 9^2,\quad
e_7 = 2 \times 9^3,\;\dots
\]
So for odd \(k = 2n+1\),
\[
e_{2n+1} = 2 \cdot 9^n.
\]
Since \(n = \frac{k-1}{2}\), we can write \(e_k = 2 \cdot 9^{\tfrac{k-1}{2}}\) when \(k\) is odd.
