已知等差数列{an},设{bn}=1/2的an方 已知b1+b2+b3=21/8,b1b2b3=1/8,求{an}
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设a2 = a ，a1 = a - d ，a3 = a + d
b1 * b2 * b3 = (1/2)^(a + a - d + a + d) = (1/2)^(3a) = 1/8 -> a = 1
b1 + b2 + b3 = (1/2)^(1 - d) + 1/2 + (1/2)^(1 + d) = 1/2 * [(1/2)^(-d) + 1 + (1/2)^d] = 21/8
(1/2)^d = 4 , d = -2 或 (1/2)^d = 1/4 , d = 2
an = 2n - 3 或 an = 5 - 2n
b1 * b2 * b3 = (1/2)^(a + a - d + a + d) = (1/2)^(3a) = 1/8 -> a = 1
b1 + b2 + b3 = (1/2)^(1 - d) + 1/2 + (1/2)^(1 + d) = 1/2 * [(1/2)^(-d) + 1 + (1/2)^d] = 21/8
(1/2)^d = 4 , d = -2 或 (1/2)^d = 1/4 , d = 2
an = 2n - 3 或 an = 5 - 2n
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