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04 Aug 2025, 07:23
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\(a_3\): -2
\(a_5\): 128
The infinite sequence of nonzero integers \(a_1\), \(a_2\), \(a_3\), ... is such that \(a_n = (a_{n-1})(a_{n-2})^3\) for \(n > 2\), and \(a_2 = 2\).
Select for \(a_3\) and \(a_5\) the values of \(a_3\) and \(a_5\), respectively, that would be jointly consistent with the given information. Make only two selections, one in each column.
Select for \(a_3\) and \(a_5\) the values of \(a_3\) and \(a_5\), respectively, that would be jointly consistent with the given information. Make only two selections, one in each column.
| \(a_3\) | \(a_5\) | |
| 128 | ||
| 16 | ||
| 8 | ||
| 4 | ||
| -2 | ||
| -8 |
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Official Answer
\(a_3\): -2
\(a_5\): 128
04 Aug 2025, 07:23
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Official Solution:
Given that \(a_n = (a_{n-1})(a_{n-2})^3\) and \(a_2 = 2\), we have:
\(a_{3} = (a_{2})(a_{1})^3 = 2(a_{1})^3\)
\(a_{4} = (a_{3})(a_{2})^3 = 8(a_{3})\)
\(a_{5} = (a_{4})(a_{3})^3\)
Substituting \(a_{4} = 8(a_{3})\) into \(a_{5} = (a_{4})(a_{3})^3\), we get:
\(a_{5} = 8(a_{3})(a_{3})^3 = 8(a_{3})^4\)
So, we need such a pair of \(a_3\) and \(a_5\) that \(a_{5} = 8(a_{3})^4\). Checking the options, only \(a_3 = -2\) and \(a_5 = 128\) work.
Correct answer:
\(a_3\) "-2"
\(a_5\) "128"
Bunuel
Given that \(a_n = (a_{n-1})(a_{n-2})^3\) and \(a_2 = 2\), we have:
\(a_{3} = (a_{2})(a_{1})^3 = 2(a_{1})^3\)
\(a_{4} = (a_{3})(a_{2})^3 = 8(a_{3})\)
\(a_{5} = (a_{4})(a_{3})^3\)
Substituting \(a_{4} = 8(a_{3})\) into \(a_{5} = (a_{4})(a_{3})^3\), we get:
\(a_{5} = 8(a_{3})(a_{3})^3 = 8(a_{3})^4\)
So, we need such a pair of \(a_3\) and \(a_5\) that \(a_{5} = 8(a_{3})^4\). Checking the options, only \(a_3 = -2\) and \(a_5 = 128\) work.
Correct answer:
\(a_3\) "-2"
\(a_5\) "128"
