Answer
See explanation
Work Step by Step
We have the recurrence
\[
d_k \;=\; 4\,d_{k-2},
\quad \text{for all integers }k \ge 2,
\quad\text{with}\quad d_0=1,\;d_1=-1.
\]
Because the update involves \(k-2\) rather than \(k-1\), the sequence effectively “jumps” in steps of 2. Let us compute the first few terms to spot the pattern:
\(d_0 = 1\) (given),
\(d_1 = -1\) (given),
\(d_2 = 4\,d_0 = 4 \cdot 1 = 4,\)
\(d_3 = 4\,d_1 = 4 \cdot (-1) = -4,\)
\(d_4 = 4\,d_2 = 4 \cdot 4 = 16,\)
\(d_5 = 4\,d_3 = 4 \cdot (-4) = -16,\)
\(d_6 = 4\,d_4 = 4 \cdot 16 = 64,\)
\(d_7 = 4\,d_5 = 4 \cdot (-16) = -64,\)
- and so on.
We see two interleaved subsequences:
- **Even indices** \(k=0,2,4,6,\dots\):
\(d_0=1,\; d_2=4,\; d_4=16,\; d_6=64,\dots\) which are powers of 4.
But \(4^n = 2^{2n}\). In fact, when \(k\) is even, say \(k=2n\), we get
\[
d_{2n} \;=\; 4^n \;=\; 2^{2n} \;=\; 2^k.
\]
- **Odd indices** \(k=1,3,5,7,\dots\):
\(d_1=-1,\; d_3=-4,\; d_5=-16,\; d_7=-64,\dots\) which are the negatives of the even‐index terms.
When \(k\) is odd, say \(k=2n+1\), we get
\[
d_{2n+1}
\;=\; -\,4^n
\;=\; -\,2^{2n}
\;=\; -\,2^{\,k-1}.
\]
Putting this all together, a convenient way to write the explicit solution is by cases:
\[
\boxed{
d_k
= \begin{cases}
2^k,
& \text{if \(k\) is even},\\[6pt]
-\,2^{\,k-1},
& \text{if \(k\) is odd}.
\end{cases}
}
\]
Either of these two forms captures the behavior completely:
- **Even \(k\):** \(d_k = 2^k.\)
- **Odd \(k\):** \(d_k = -\,2^{\,k-1}.\)
This satisfies both the recurrence \(d_k = 4\,d_{k-2}\) and the initial conditions \(d_0=1,\;d_1=-1.\)
