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Infinitely many mathematicians walk into a bar. The first says "I'll have a beer". The next ones say "I'll have half of the previous guy". The bartender pours out 2 beers and says "Know your limits".

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A sequence adheres to the recursion relation \(a_{n}=\sqrt[3]{a_{n-1}a_{n-2}a_{n-3}}\) with initial terms \(a_{0}=1,a_{1}=2,a_{2}=1\).

As \(n\) tends to infinity, what is the limit of \( a_n\) correct to 2 decimal places?

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Evaluate

\[ \sqrt{\dfrac{1}{2}}\times\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2}}}\times\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2}}}}\cdots\]

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The sequence \(1, 2, 3, 2, \ldots\) has the property that every fourth term is the average of the previous three terms. That is, the sequence \(\{a_n\}\) is defined as \(a_1 = 1, a_2 = 2, a_3 = 3\), and \[a_{n+3} = \dfrac{a_{n+2} + a_{n+1} + a_n}{3}\] for any positive integer \(n\).

This sequence converges to a rational number \(\frac{a}{b}\), where \(a\) and \(b\) are coprime positive integers. Find \(a+b\).

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