Question

已知数列｛an｝，a1=3/5 ,a2=31/100,且数列｛an+1-1/10an}是公比为1/2的等比数列，数列｛lg(an+1-1/2an)是

Answer 1

解:
∵{a(n+1)-[an/10]}是公比为1/2的等比数列bn
∴b1=a2-[a1/10]=(31/100)-(3/50)=1/4
bn=(1/4)×(1/2)＾(n-1)=(1/2)＾(n+1)
bn=a(n+1)-(an/10)=(1/2)＾(n+1)
∴a(n+1)-(an/10)=(1/2)＾(n+1)........(1)
数列Cn={lg(a(n+1)-an/2)}是公差为-1的等差数列
C1=lg[(a2)-(a1/2)]
=lg(1/100)=-2
d=-1
Cn=-2+(n-1)×(-1)=-n-1=lg[a(n+1)-an/2]
∴(1/10)＾(n+1)=a(n+1)-(an/2)........(2)
(1)-(2):
(4/10)an=(1/2)＾(n+1)-(1/10)＾(n+1)
an=(5/2)[(1/2)＾(n+1)-(1/10)＾(n+1)]