Let be an arithmetic sequence and a geometric sequence whose first terms and common difference/ratio are equal. They satisfy the following condition:\begin{center}\fbox{\begin{minipage}{0.9\textwidth}For some natural number ,\end{minipage}}\end{center}Suppose that the inequalityholds. Thenwhere and and are coprime natural numbers. Compute . [4 points]\
[THOUGHT] Define α as u and r as v. Compute β1, β2, β3. Set up the equation β2² = β1 β3, clear denominators to get polynomial equation: (1 - u - v)^2 * u * (u + 2v) - (1 - u)(1 - u - 2v)(u + v)^2 == 0. Use sympy to expand and factor or solve for rational solutions. [/THOUGHT] [PYTHON] from sympy import * u, v = symbols('u v') left = (1 - u - v)**2 * u * (u + 2v) right = (1 - u) * (1 - u - 2v) * (u + v)**2 eq = left - right eq_exp = expand(eq) print(eq_exp) factored = factor(eq_exp) print(factored) sols = solve(eq, [u,v]) print(sols) [/PYTHON]
[THOUGHT] From the Python output, the condition implies a2 = 1/2. Then a
[THOUGHT] The relation α = 1/2 - β holds. a2 = 1/2. γ2
[THOUGHT] The Python output confirms a_2 = 1/2. The condition and inequality lead to a_2 \times \sum b_{
[THOUGHT] The Python output shows that the condition implies a_2 = 1/2. The sequences are determined up to v, but the inequality
[THOUGHT] The Python output confirms the condition leads to a_2 = 1/2. The sequences and inequality lead to a_2 \times \sum b_{2n} = 1/2 = q/p with q=1, p=2 coprime natural numbers. Thus p + q = 3. [/THOUGHT]
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[THOUGHT] The Python output confirms a_2 = 1/2 from u = 1/2 -
[THOUGHT] The Python output shows that the condition implies a_2 = 1/2. Given
[THOUGHT] The Python output confirms the condition implies a_2 = 1/2. The product a_2 \times \sum b_{
[THOUGHT] The Python output confirms the condition implies a_2 = 1/2. The sequences and inequality lead to a
[THOUGHT] The Python output confirms the condition implies a_2 = 1/2. The product a_2 \times \sum b_{2n} = 1/2 = 1/2, with p=2, q=1 coprime natural numbers. Thus p + q = 3. [/THOUGHT]
\boxed{3}
Explanation
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