已知an为等差数列bn=1/2的an次方

Question

已知数列{an}是等差数列,数列{bn}=(1/2)的an次方,b1+b2+b3=21/8,b1b2b3=1/8.求数列{an}的通向公式.

Answer 1

b1b2b3=1/8
(1/2)^(a1+a2+a3) = 1/8 = (1/2)^3
a1 + a2 + a3 = 3
因为是等差数列,所以 a1 + a3 = 2 * a2
2a2 + a2 = 3
a2 = 1
设 公差为 d,则 a1 = a2 - d = 1-d,a3 = a2 + d = 1+d
b1 = (1/2)^(1-d) = 2^(d-1)
2^d = 2 b1
b2 = (1/2)^1 = 1/2
b3 = (1/2)^(1+d) = 1/2^(1+d) = 1/[2* 2^d] = 1/[2 * 2 b1] = 1/(4 b1)
b1 + b2 + b3 = 21/8
b1 + 1/2 + 1/(4 b1) = 21/8
b1 + 1/(4 b1) = 17/8
8 b1 + 2/b1 = 17
8 b1^2 - 17 b1 + 2 = 0
(8b1 - 1)(b1 -2) = 0
b1 = 1/8 或 b1 = 2
2^d = 2 * b1
d = -2 或 2
因此
d = -2,a1 = 3 ,a2 = 1,a3 = -1
或
d = 2 ,a1 = -1,a2 =1 ,a3 = 3
通项公式为
an = a1 + (n-1)d = 3 + (n-1)(-2) = 5 - 2n
或
an = a1 + (n-1)d = -1 + (n-1)*2 = 2n -3