The output of the circuit given as $: Q=aQ_{n-1}+ab+bQ_{n-1}$
Hence, $Q_{n}=Q_{n-1}(a+b)+ab$
$00 \implies Q_{n-1}(0+0) + 0.0 = Q_{n-1}(0) + 0 = 0+0 = 0$
$01 \implies Q_{n-1}(0+1) + 0.1= Q_{n-1} (1)+ 0 = Q_{n-1}+0 = Q_{n-1}$
$10 \implies Q_{n-1}(1+0) + 1.0 = Q_{n-1} (1) + 0 = Q_{n-1}+0= Q_{n-1}$
$11 \implies Q_{n-1}(1+1)+ 1.1 =Q_{n-1}(1) + 1 =Q_{n-1}+1 = 1$
$${\begin{array}{cc|c}
\textbf{a}& \textbf{b}& \bf{ Q_n} \\\hline
0&0&0\\ 0&1& Q_{n-1} \\ 1&0& Q_{n-1} \\ 1&1 &1\\
\end{array}}$$