finding the output of these three systems when connected in series?
You are right and the manual is wrong.

Given $S_1, S_2, S_3$ and the respective input-output signals as below :

$$\\x[n] \rightarrow \boxed{S_1} \rightarrow v[n] \rightarrow \boxed{S_2} \rightarrow w[n] \rightarrow \boxed{S_3}\rightarrow y[n] \\$$

$\\$

$ x[n] \xrightarrow{S1} v[n] = \begin{cases}
x[n/2] &amp; \text{if n is even;}\\
0 &amp; \text{if n is odd.}\\
\end{cases}$

$ v[n] \xrightarrow{S2} w[n] = v[n] + (1/2)v[n-1] + (1/4)v[n-2] $

$ w[n] \xrightarrow{S3} y[n] = w[2n] \\ \\$

You would compute the output $y[n]$ in three steps as: (you have compined step 1 into 2 actually)

$v[n] \xrightarrow{S2} w[n] $

$
\begin{align}
w[n] &amp;= v[n] + (1/2)v[n-1] + (1/4)v[n-2]  \\ \\
&amp;=  \begin{cases} {
x[n/2] + (1/4)x[(n-2)/2] ~~~~~~ \text{if n is even;}\\
(1/2)x[(n-1)/2] ~~~~~~~~~~~~~~~~~~~~~~ \text {if n is odd.} }
\end{cases}
\end{align}
$

$\\$

and then, feeding $w[n]$ to $S_3$, we have:

$w[n] \xrightarrow{S3} y[n] = w[2n] \\$

$ y[n] =  \begin{cases}
x[2n/2] + (1/4)x[(2n-2)/2] &amp; \text{if 2n is even;}\\
(1/2)x[(2n-1)/2] &amp; \text {if 2n is odd.}\\
\end{cases}$

Hence (your) the solution is right: 
$$y[n] = x[n] + (1/4)x[n-1]. $$

solution manual is wrong: 
$$x[n] \xrightarrow{S1,S2,S3} y[n] = x[n] + (1/2)x[n-1] + (1/4)x[n-2].$$ 

You can verify this simply by putting some test signals, or even easier by writing a simple MATLAB/Octave/Python script to implement the $S_1,S_2,S_3$ system.