<s>[INST] Finding the general formula for the sequence with $d_0=1$, $d_1=-1$, and $d_k=4 d_{k-2}$ 
Suppose that we want to find a general formula for the terms of the sequence
$$d_k=4 d_{k-2}, \text{ where } d_0=1 \text{ and } d_1=-1$$

I have done the following:
\begin{align*}d_k=4d_{k-2}&=2^2d_{k-2} \\ &=2^2\left (2^2d_{(k-2)-2}\right )=2^4d_{k-4} \\ &  =2^4\left (2^2d_{(k-4)-2}\right )=2^6d_{k-6} \\ & = 2^6\left (2^2d_{(k-6)-2}\right )=2^8d_{k-8} \\ & = \ldots \\ & = 2^id_{k-i}\ , \ \ i \text{ even}\end{align*}
If $k$ even, then at the last step we have for $i=k$ (since $k$ is the maximum even number  $\leq k$) : $d_k=2^kd_{k-k}=2^kd_0=2^k$.
If $k$ odd, then at the last step we have for $i=k-1$ (since $k-1$ is the maximum even number $\leq k$) : $d_k=2^{k-1}d_{k-(k-1)}=2^{k-1}d_1=-2^{k-1}$.
How can we find the general form for the terms of the recurrence relation? Or do we distinguish cases when $k$ is even and odd?
I am interested to find the general formula without using the characteristic equation.
 [/INST] You found $d_k=2^k$ when $k$ is even and $d_k=-2^{k-1}$ when $k$ is odd.
To put this in one formula, note that $\dfrac{1+(-1)^n}2$ is $0$ when $n$ is odd and $1$ when $n$ is even, 
whereas $\dfrac{1-(-1)^n}2$ is $1$ when $n$ is odd and $0$ when $n$ is even.
So you could say $d_k=2^k\dfrac{1+(-1)^k}2-2^{k-1}\dfrac{1-(-1)^k}2,$ 
which simplifies to $2^{k-2}\left[1+3(-1)^k\right]$.
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