In the given network of AND and OR gates,the | vedclass

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(D) Let us trace the output of the logic gates step by step.Step $1$: The first $AND$ gate takes inputs $X_0$ and $X_1$,giving output $Y_1 = X_0 X_1$.

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$X_0 X_1 \ldots X_{n-1}+X_2 X_3 X_5 \ldots X_{n-1}+X_{n-2} X_{n-1}+X_n$

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(D) Let us trace the output of the logic gates step by step.Step $1$: The first $AND$ gate takes inputs $X_0$ and $X_1$,giving output $Y_1 = X_0 X_1$.Step $2$: The first $OR$ gate takes $Y_1$ and $X_2$,giving output $Y_2 = X_0 X_1 + X_2$.Step $3$: The next $AND$ gate takes $Y_2$ and $X_3$,giving output $Y_3 = (X_0 X_1 + X_2) X_3 = X_0 X_1 X_3 + X_2 X_3$.Step $4$: The next $OR$ gate takes $Y_3$ and $X_4$,giving output $Y_4 = X_0 X_1 X_3 + X_2 X_3 + X_4$.Continuing this pattern,for an even $n$,the final output $Q$ will be of the form: $Q = X_0 X_1 X_3 X_5 \ldots X_{n-1} + X_2 X_3 X_5 \ldots X_{n-1} + X_4 X_5 \ldots X_{n-1} + \ldots + X_{n-2} X_{n-1} + X_n$.

In the given network of $AND$ and $OR$ gates,the output $Q$ can be written as (assuming $n$ is even):

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