

(Ⅰ)求{an}的通项公式;
(Ⅱ)若数列{bn}满足b1=1,且bn+1=bn+an,求数列{bn}的通项公式.
(Ⅰ)求{an}的通项公式;
(Ⅱ)若数列{bn}满足b1=1,且bn+1=bn+an,求数列{bn}的通项公式.
解:
(Ⅰ)依题意得:a2a4=(又∵=a2a4,a3>0.∴a3=
∵a2+a3=2×,∴a2=
,
故公比q=,a1=1. an=a1qn-1=21-n,
即{an}的通项公式为an=a1qn-1=21-n
(Ⅱ)∵bn+1=bn+an(n=1,2,3,…),
∴bn+1-bn=()n-1
得b2- b1=1,
b3-b2=,
b4-b3=()2,
…,
bn-bn-1=()n-2
各式相加得:bn- b1=1++(
)2+(
)3+…+(
)n-2
==2-2(
)n-1
又∵b1=1,∴bn=3-()n-2