Finding Fourier series coefficients for discrete time signal
You are asking for DFS (discrete Fourier series) coefficients $b_k$ of the periodic sequence $y[n]$, in terms $a_k$ of the periodic sequence $x[n]$. Since DFS and DFT (discrete Fourier transform) are the same things, I will instead write down an answer for the DFTs $Y[k]$ and $X[k]$ of the associated sequences, in the context of which $X[k] = a_k$ , and $Y[k] = b_k$ will be understood.
I would like to relate DFTs to DTFTs (discrete-time Fourier transform) through the sampling relation, which indicates that :

N-point DFT X[k] of the N-point sequence x[n], is the uniform samples of the DTFT X(w) of
x[n].

The DTFT of $x[n]$ is :
$$ X(\omega) = \sum_{n=0}^{N-1} x[n] e^{-j \omega n } \tag{a}$$
and the DFT of $x[n]$ is :
$$ X[k] = X(\frac{2\pi}{N}k) = \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N}nk } \tag{b}$$
Let $x[n]$ and $y[n]$ be two sequences related by:
$$y[n]=\begin{cases}{ x[n] ~~~,~~~n:even \\  ~~~0 ~~~~~,~~~ n:odd }\end{cases}$$
You can also express $y[n]$ as :
$$ x[n] \longrightarrow \boxed{ \downarrow 2 } \overset{w[n]}{\longrightarrow} \boxed{ \uparrow 2 }    \longrightarrow y[n] $$
Applying DTFT relations to the sequences :
$$ W(\omega) = 0.5 \left( X\left( \frac{\omega}{2} \right) +  X\left( \frac{\omega-2\pi}{2} \right) \right) \tag{1} $$
and
$$ Y(\omega) = W(2\omega) \tag{2} $$
hence
$$ Y(\omega) = 0.5 \left( X\left( \omega \right) +  X\left( \omega-\pi\right) \right) \tag{3} $$
Eq.3 describes the relation (for any $N$) between the DTFTs $Y(\omega)$ and $X(\omega)$ of the sequences $x[n]$, and $y[n]$. To turn this into the equivalent relation between the N-point DFTs $X[k]$ and $Y[k]$, (and for your case a relation between DFS  $a_k$ and $b_k$) we shall sample the associated DTFTs. Then we have :
$$  \begin{align}
Y[k] &amp;= Y( \frac{2\pi}{N} k ) ~~~,~~~k = 0,1,...,N-1 \tag{4} \\ \\
     &amp;= 0.5 \left( X\left( \frac{2\pi}{N} k \right) +  X\left( \frac{2\pi}{N} k-\pi\right) \right) \tag{5} \\ \\
     &amp;\boxed{ Y[k] = 0.5 \left( X\left[ k \right] +  X\left[k- \frac{N}{2} \right] \right) }\tag{6} \\
\end{align}$$
Eq.6 describes the requested relation between N-point DFTs $X[k]$ and $Y[k]$.
When $N$ is EVEN (say $N = 2 M$ ) the relation becomes :
$$ \boxed{ Y[k] = 0.5 ( X[k] + X[k-M] ) }\tag{7} $$
which is what you have arrived at $b_k = 0.5(a_k + a_{k -\frac{N}{2}})$.
However, when $N$ is ODD, there is a problem due to $N/2$ not being an integer. The sequence $X[k-N/2]$ is interpreted as an interpolation of $X[k]$ at the fractional indices $k-N/2$.
For $N=2M+1$ , we have $X[k-N/2] = X[k-M-0.5]$ , $k=0,1,...,N-1$ , which evaluates $X[k]$ at the points $m = 0.5,1.5,...,N-0.5$. To get those intermediate samples, we need an interpolation of $X[k]$ by 2.
Interpolation of $X[k]$ by 2, is achieved by zero padding $x[n]$ by $N$ samples, and computing its $2N$-point DFT :
$$ x_e[n] = \begin{cases}{ x[n] ~~~,~~~ 0 \leq n &lt; N \\ ~~~0 ~~~~ , ~~~~ N \leq n &lt; 2N }\end{cases} $$
Then we can see, by the DFT - DTFT sampling relation that :
$$ \mathcal{DFT}\{x_e[n]\} =X_2[k] = X(\frac{2\pi}{2N}k )= X(\frac{2\pi}{N} k/2 ) = X[k/2] \tag{8} $$
for $ k = 0,1,.,2N-1$. Note that DTFTs of $x[n]$ and $x_e[n]$ are the same.
Based on Eqs. 8, you can restate Eq.6 when $N$ is odd :
$$ \boxed{ Y[k]= 0.5 \big( X\left[ k \right] +  X_2\left[2k- N\right] \big) } \tag{9} $$
with your notation it becomes:
$$ b_k = 0.5 ( a_k + a^e_{2k-N} ) $$ where $a^e$ is the DFS of the sequence $x_e[n]$.
As you can see, you cannot express $b_k$ (with a simple formula) in terms of sole $a_k$ when $N$ is odd.