若{an}为等差数列,bn=(1/2)^an,b3b4b5=512,b1b5+3b2b6+b3b7=464,求an若{an}为等差数列,bn=(1/2)^an,b3b4b5=512,b1b5+3b2b6+b3b7=464,求{an}的通项公式.

Question

若{an}为等差数列,bn=(1/2)^an,b3b4b5=512,b1b5+3b2b6+b3b7=464,求an
若{an}为等差数列,bn=(1/2)^an,b3b4b5=512,b1b5+3b2b6+b3b7=464,求{an}的通项公式.

Answer 1

B(n+1)/Bn=(1/2)^A(n+1)/(1/2)^An=(1/2)^(A(n+1)-An)=(1/2)^d
Bn是等比数列
B3×B5=(B4)^2
B3×B4×B5=(B4)^3=512
B4=8
B2×B6=(B4)^2=64
B1×B5=(B2/q)×(B5/q)=64/(q^2)
B3×B7=B2×q×B5×q=64×q^2
64/(q^2)+3×64+64×q^2=464
(1/q^2)+(q^2)=17/4
q=±2
B4=(1/2)^A4=8
A4=-3
当q=-2时,
q=B(n+1)/Bn=(1/2)^d>0≠-2
舍去
当q=2时,
q=B(n+1)/Bn=(1/2)^d=2
d=-1
A1=A4-3d=-3-3(-1)=0
An=A1+(n-1)d=1-n