(1/2)设an是等差数列,bn=1/2的an次方,已知b1+b2+b3=21/8,b1·b2·b3=1/8,求等差数列的通项an n在字母

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(1/2)设an是等差数列,bn=1/2的an次方,已知b1+b2+b3=21/8,b1·b2·b3=1/8,求等差数列的通项ann在字母(1/2)设an是等差数列,bn=1/2的an次方,已知b1+

(1/2)设an是等差数列,bn=1/2的an次方,已知b1+b2+b3=21/8,b1·b2·b3=1/8,求等差数列的通项an n在字母
(1/2)设an是等差数列,bn=1/2的an次方,已知b1+b2+b3=21/8,b1·b2·b3=1/8,求等差数列的通项an n在字母

(1/2)设an是等差数列,bn=1/2的an次方,已知b1+b2+b3=21/8,b1·b2·b3=1/8,求等差数列的通项an n在字母
b1·b2·b3=1/8 即 (1/2)^(a1+a2+a3)=1/8
所以 a1+a2+a3=3
因为an为等差数列所以3*a2=3 a2=1
设公差为d
因为b1+b2+b3=21/8
所以(1/2)^(1-d)+1/2+(1/2)^(1+d)=21/8
解得d=2
所以an的通项公式为an=2n-3

bn=(1/2)^(an)
b1b2b3=(1/2)(a1+a1+d+a1+2d)=(1/2)(3a1+3d)=(1/8)^(a1+d)=1/8
a1+d=1 a2=1
b1+b2+b3=(1/2)^(a2-d)+(1/2)^(a2)+(1/2)^(a2+d)
=(1/2)^(a2)[(1/2)^(-d)+1+(1/2)^d]
=(1/2)[(1/2)^(...

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bn=(1/2)^(an)
b1b2b3=(1/2)(a1+a1+d+a1+2d)=(1/2)(3a1+3d)=(1/8)^(a1+d)=1/8
a1+d=1 a2=1
b1+b2+b3=(1/2)^(a2-d)+(1/2)^(a2)+(1/2)^(a2+d)
=(1/2)^(a2)[(1/2)^(-d)+1+(1/2)^d]
=(1/2)[(1/2)^(-d)+1+(1/2)^d]=21/8
(1/2)^(-d)+1+(1/2)^d=21/4
(1/2)^d-17/4+(1/2)^(-d)=0
等式两边同乘以4×(1/2)^d
4×[(1/2)^d]^2-17×(1/2)^d+4=0
[(1/2)^d-4][4×(1/2)^d-1]=0
(1/2)^d=4 d=-2或(1/2)^d=1/4 d=2
d=-2时，a1=a2-d=1-(-2)=3 an=a1+(n-1)d=3+(-2)(n-1)=5-2n
d=2时，a1=a2-d=1-2=-1 an=a1+(n-1)d=-1+2(n-1)=2n-3

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由题意，知b1=(1/2)^a1>0,当n≥2时，bn/bn-1=(1/2)^(an-an-1)=(1/2)^d为常数，故{bn}是等比数列，设（1/2)^d=q,由b1·b2·b3=1/8得b2=1/2,所以1/2q+1/2+q/2=21/8,解得q=1/4或q=4
当q=1/4时，（1/2)^d=1/4,所以d=2 , 又b1=(1/2)^a1=2,所以a1=-1,于是，an=2n-3...

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令an的公差为d
b1*b2*b3=1/2^a1*1/2^a2*1/2^a3
=1/2^a1*1/2^(a1+d)*1/2^(a1+2d)
=2^(-3a1-3d)=1/8
-3a1-3d=-3
a1+d=1
b1+b2+b3=1/2^a1+1/2^a2+1/2^a3
=1/2^a1+1/2^(a1+d)+1/2^(a1+2d)
=1...

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令an的公差为d
b1*b2*b3=1/2^a1*1/2^a2*1/2^a3
=1/2^a1*1/2^(a1+d)*1/2^(a1+2d)
=2^(-3a1-3d)=1/8
-3a1-3d=-3
a1+d=1
b1+b2+b3=1/2^a1+1/2^a2+1/2^a3
=1/2^a1+1/2^(a1+d)+1/2^(a1+2d)
=1/2^a1+1/2^(a1+1-a1)+1/2^(a1+2-2a1)
=1/2^a1+1/2+1/2^(2-a1)=21/8
1/2^a1+1/2^(2-a1)=17/8
2^(2-a1)+2^a1=17/2
a1=-1,d=2
an=a1+(n-1)d=-1+2(n-1)=2n-3

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设等差数列an首项a1,公差d
b1·b2·b3=1/8 (1/2)^3(a1+d)=1/8 a1+d=1
b1+b2+b3=21/8 (1/2)^a1[1+(1/2)^d+(1/2)^2d]=21/8
解得：a1= -1 d=2 或 a1=3 d= -2
an =a1+(n-1)d=2n-3 或5-2n
故：等差数列的通项an =2n-3 或5-2n

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