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04 Jul 2025, 04:18
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\(a_1, a_2, a_3\)..... are all nonzero integers and are defined \(a_n = (a_{n-1})(a_{n-2})^3\) where \(n>2\) and \(a_2=2\)
\(a_3=a_2(a_1)^3\)
\(a_3=2(a_1)^3\)
\(a_4=a_3(a_2)^3\)
\(a_4 = a_2(a_1)^3(2)^3\)
\(a_4 =16 (a_1)^3\)
We have \((a_1)^3\) in both \(a_3\) and \(a_4\). Find \((a_1)^3\) in terms of \(a_3\) and \(a_4\) and set them equal to eachother
\((a_1)^3 = \frac{a_3}{2}\)
\((a_1)^3 = \frac{a_4}{16}\)
\(\frac{a_3}{2} = \frac{a_4}{16}\)
\(a_4 = 8a_3\)
\(a_5 = a_4(a_3)^3\)
\(a_5 = 8a_3(a_3)^3\)
\(a_5 = 8(a_3)^4\)
When \(a_3 = -2\), \(a_5 = 128\)
\(a_3=a_2(a_1)^3\)
\(a_3=2(a_1)^3\)
\(a_4=a_3(a_2)^3\)
\(a_4 = a_2(a_1)^3(2)^3\)
\(a_4 =16 (a_1)^3\)
We have \((a_1)^3\) in both \(a_3\) and \(a_4\). Find \((a_1)^3\) in terms of \(a_3\) and \(a_4\) and set them equal to eachother
\((a_1)^3 = \frac{a_3}{2}\)
\((a_1)^3 = \frac{a_4}{16}\)
\(\frac{a_3}{2} = \frac{a_4}{16}\)
\(a_4 = 8a_3\)
\(a_5 = a_4(a_3)^3\)
\(a_5 = 8a_3(a_3)^3\)
\(a_5 = 8(a_3)^4\)
When \(a_3 = -2\), \(a_5 = 128\)
04 Jul 2025, 04:36
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Let a1 = x, a2=2
Then:
a3 = a2 * (a1)^3 = 2x^3
a4 = a3 * (a2)^3 = (2x)^3 * 8 = 16x^3
a5 = a4 * (a3)^3 = (16*x^3)(2*x^3)^3 = 16*x^3 * 8*x^9 = 128 * x^12
=> a5 = 128*x^12
x^12 is larger or equal to 0 => a^5 is larger or equal 128 => a^5 = 128(among options in column)
=> a1 may be -1 or 1
=> a3 maybe -2 or 2.
There's no 2 as an option, so a3 = -2
Answer: a3 = -2, a5 = 128
Then:
a3 = a2 * (a1)^3 = 2x^3
a4 = a3 * (a2)^3 = (2x)^3 * 8 = 16x^3
a5 = a4 * (a3)^3 = (16*x^3)(2*x^3)^3 = 16*x^3 * 8*x^9 = 128 * x^12
=> a5 = 128*x^12
x^12 is larger or equal to 0 => a^5 is larger or equal 128 => a^5 = 128(among options in column)
=> a1 may be -1 or 1
=> a3 maybe -2 or 2.
There's no 2 as an option, so a3 = -2
Answer: a3 = -2, a5 = 128
04 Jul 2025, 04:40
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recursive rule: a_n =(a_n-1)(a_n-2)^3 and a_2 = 2
Using the rule, we express a_(3) = 2.a_1 ^ 3 and a_(5) = 8a_(3) ^ 4
Testing a_(1) = - 1
a_(3) = - 2.
Plugging into the formula, a_(5) = 8 * (- 2) ^ 4 = 8×16 128.
Thus,a_(3) = - 2 and a_(5) = 128.
Using the rule, we express a_(3) = 2.a_1 ^ 3 and a_(5) = 8a_(3) ^ 4
Testing a_(1) = - 1
a_(3) = - 2.
Plugging into the formula, a_(5) = 8 * (- 2) ^ 4 = 8×16 128.
Thus,a_(3) = - 2 and a_(5) = 128.
