Discussion
Using Boolean equation, the output 'Y' of the network shown below is equal to :
Answer: D
The output is \( Y = X_{0}X_{1}X_{3}\ldots X_{n-1} + X_{2}X_{3}X_{5}\ldots X_{n-1} \) \( + \ldots + X_{n-2}X_{n-1} + X_{n} \).
The gates alternate AND and OR. Work along the chain:
\[ \begin{aligned} G_{1} &= X_{0}X_{1} \\ G_{2} &= X_{0}X_{1} + X_{2} \\ G_{3} &= X_{0}X_{1}X_{3} + X_{2}X_{3} \end{aligned} \]
Each AND gate multiplies every existing term by its new input, and each OR gate adds its input as a new term. Every term therefore carries all the AND-gate inputs after it (\( X_{3}, X_{5}, \ldots, X_{n-1} \)), and the final OR adds \( X_{n} \) alone.
All-OR, \( X_{0} + X_{1} + \ldots + X_{n} \), would need every gate to be an OR. The sum of pairs ignores the cascading, and the first option multiplies in inputs such as \( X_{2} \) and \( X_{n} \) that enter through OR gates.
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