Question

已知等差数列{an}，设bn=（1/2）an次幂，又已知b1+b2+b3=21/8,b1乘b2乘b3=1/8 求{an}的通项公式

Answer 1

因为an是等差数列，设公差是d，首项a1，
那么有
an-a(n-1)=d

bn=(1/2)^(an）
b(n-1)=(1/2)^(a(n-1))
=>bn/b(n-1)=(1/2)^(an-a(n-1))=(1/2)^d
因此bn是等比数列，公比q=（1/2)^d

b1=(1/2)^a1
b1*b2*b3=b2^3=1/8
=>b2=1/2

又
b1+b2+b3=b2(1/q+1+q)=1/2*(1/q+1+q)=21/8
=>q=4或1/4
有q=(1/2)^d
=》d=-2或2

b2=b1*(1/2)^d=1/2
=>b1=8或1/8

b1=(1/2)^a1=8
=>a1=-3或3

因此an=a1+(n-1)d=-3+(n-1)*(-2)=-1-2n
或an=3+(n-1)*2=2n+1

Answer 2

已知等差数列an.设bn=(1/2)^an, 已知b1+b2+b3=21/8b1*b2*b3=1/8求an的通项公式。解:b1+b2+b3=21/8,→(1/2)^a1+(1/2)^a2+(1/2)^a3=21/8,....(1) b1*b2*b3=1/8,→,→(1/2)^a1*(1/2)^a2*(1/2)^a3=1/8,→ (1/2)^(a1+a2+a3)=(1/2)^3,→a1+a2+a3=3, 等差数列an:a1+a3=2a2,→a2=1,a1+a3=2,a3=2-a1,代入(1): (1/2)^a1+(1/2)^1+(1/2)^(2-a1)=21/8,→ (1/2)^a1+1/2+(1/2)^(2-a1)=21/8,→ (1/2)^a1+[(1/2)^2]/[(1/2)^a1]=17/8→ (1/2)^a1+1/4[(1/2)^a1]=17/8→ 4[(1/2)^a1]^2+1=17/2[(1/2)^a1]→ 8[(1/2)^a1]^2-17[(1/2)^a1]+2=0→ [8(1/2)^a1-1]^[(1/2)^a1-2]^=0→ (1/2)^a1=1/8,或(1/2)^a1=2,→ a1=3或a1=-1,→ a1=3,a2=1,a3=-1,或 a1=-1,a2=1,a3=3 ∴(1)an=3+(n-1)*(-2),即an=-2n+5,或 (2)an=-1+(n-1)*2,即an=2n-3

Answer 3

http://zhidao.baidu.com/question/179725857.html?si=1

Answer 4

很好