Question

Which one of the following options is correct for the given data in the table? \[ \begin{array}{|c|c|c|c|c|} \hline \text{Iteration (i)} & 0 & 1 & 2 & 3
\hline \text{Input (I)} & 20 & -4 & 10 & 15
\hline \text{Output (X)} & 20 & 16 & 26 & 41
\hline \text{Output (Y)} & 20 & -80 & -800 & -12000
\hline \end{array} \]

• A. \( X(i) = X(i-1) + I(i);\; Y(i) = Y(i-1)I(i); \;\; i>0 \)
• B. \( X(i) = X(i-1)I(i);\; Y(i) = Y(i-1) + I(i); \;\; i>0 \)
• C. \( X(i) = X(i-1)I(i);\; Y(i) = Y(i-1)I(i);\; i>0 \)
• D. \( X(i) = X(i-1) + I(i);\; Y(i) = Y(i-1)(i-1);\; i>0 \)

Hint:

When solving table-driven sequence problems, test the formula step by step for each iteration. One wrong step can immediately eliminate incorrect options.

Answer 1

The Correct Option is A

Step 1: Check relation for \(X(i)\).
At \(i=0\), \(X(0)=20\). For \(i=1\): \[ X(1)=X(0)+I(1)=20+(-4)=16 \] Matches given. For \(i=2\): \[ X(2)=X(1)+I(2)=16+10=26 \] Correct. For \(i=3\): \[ X(3)=X(2)+I(3)=26+15=41 \] Matches perfectly. Thus, \(X(i)=X(i-1)+I(i)\). Step 2: Check relation for \(Y(i)\).
At \(i=0\), \(Y(0)=20\). For \(i=1\): \[ Y(1)=Y(0)\times I(1)=20 \times (-4)=-80 \] Correct. For \(i=2\): \[ Y(2)=Y(1)\times I(2)=-80 \times 10=-800 \] Correct. For \(i=3\): \[ Y(3)=Y(2)\times I(3)=-800 \times 15=-12000 \] Matches exactly. Step 3: Conclude.
Both conditions for \(X(i)\) and \(Y(i)\) match option (A). \[ \boxed{X(i)=X(i-1)+I(i), \quad Y(i)=Y(i-1)I(i)} \]