Question

Questions 43 and 44 are based on the following instructions: each question gives a statement followed by three conclusions, I, II and III. Pick the option that tells you which of the three conclusions can be derived from the statement alone.

Statement: \(A_0, A_1, A_2, \ldots\) is a sequence of numbers with \(A_0 = 1\), \(A_1 = 3\), and \(A_t = (t+1)A_{t-1} - tA_{t-2}\) for \(t = 2, 3, 4, \ldots\)

Conclusion I: \(A_8 = 77\)
Conclusion II: \(A_{10} = 121\)
Conclusion III: \(A_{12} = 145\)

• A. Using the given statement, only conclusion I can be derived.
• B. Using the given statement, only conclusion II can be derived.
• C. Using the given statement, only conclusion III can be derived.
• D. Using the given statement, none of the three conclusions I, II and III can be derived.

Hint:

Look at the difference \(A_t - A_{t-1}\) between consecutive terms rather than recomputing every term from the recurrence; that difference follows a much simpler pattern than the original recurrence does.

Answer 1

The Correct Option is D

Step 1: Find the pattern in the consecutive differences.
Let \(d_t = A_t - A_{t-1}\) be the gap between one term and the next.
We are given \(A_0 = 1\) and \(A_1 = 3\), so \(d_1 = A_1 - A_0 = 2\).
The recurrence \(A_t = (t+1)A_{t-1} - tA_{t-2}\) can be rewritten as \(A_t - A_{t-1} = t(A_{t-1} - A_{t-2})\), which means \(d_t = t \cdot d_{t-1}\).

Step 2: Turn the pattern into a closed formula for \(d_t\).
Since \(d_t = t \cdot d_{t-1}\) and \(d_1 = 2\), each step multiplies the previous gap by the next integer: \(d_2 = 2d_1\), \(d_3 = 3d_2\), and so on.
Unrolling this chain gives \(d_t = 2 \cdot t!\) for every \(t \geq 1\) (check: \(d_1 = 2 \cdot 1! = 2\), \(d_2 = 2 \cdot 2! = 4\), \(d_3 = 2 \cdot 3! = 12\), which matches direct calculation term by term).

Step 3: Add up the gaps to get a formula for \(A_t\).
\(A_t\) is just \(A_0\) plus all the gaps up to \(t\):
\[ A_t = A_0 + \sum_{k=1}^{t} d_k = 1 + 2\sum_{k=1}^{t} k! \]

Step 4: Compute \(A_8\), \(A_{10}\) and \(A_{12}\).
\(\sum_{k=1}^{8} k! = 1+2+6+24+120+720+5040+40320 = 46233\), so \(A_8 = 1 + 2(46233) = 92467\).
\(\sum_{k=1}^{10} k! = 46233 + 362880 + 3628800 = 4037913\), so \(A_{10} = 1 + 2(4037913) = 8075827\).
\(\sum_{k=1}^{12} k! = 4037913 + 39916800 + 479001600 = 522956313\), so \(A_{12} = 1 + 2(522956313) = 1045912627\).

Step 5: Compare with the three conclusions.
Conclusion I says \(A_8 = 77\), but we found \(A_8 = 92467\), so Conclusion I fails.
Conclusion II says \(A_{10} = 121\), but we found \(A_{10} = 8075827\), so Conclusion II fails.
Conclusion III says \(A_{12} = 145\), but we found \(A_{12} = 1045912627\), so Conclusion III fails.
None of the three values in the conclusions come anywhere close to the actual terms of the sequence, which grow extremely fast because of the factorial term hidden in the recurrence.

Final Answer:
Since none of the three conclusions can be derived from the statement, the correct choice is option (EE). \[ \boxed{\text{Option (EE)}} \]