Discussion
Find \(a_{12}\) if \(a_{n+1}^{2} = 5a_{n}^{2}\), where \(a_{n} \gt 0\) for \(n \ge 0\) and \(a_{0} = 2\) :
Answer: A
\(a_{12} =\) 31250.
Put \(b_{n} = a_{n}^{2}\). The recurrence becomes \(b_{n+1} = 5b_{n}\) with \(b_{0} = 4\), so \(b_{n} = 4 \cdot 5^{n}\). Since \(a_{n} \gt 0\), \(a_{n} = 2 \cdot 5^{n/2}\). Then
\[ a_{12} = 2 \cdot 5^{6} = 2 \times 15625 = 31250 \]
The positivity condition is what fixes the sign of each square root; without it \(a_{12}\) could be \(\pm 31250\). The other options are near misses with no derivation behind them.
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