等比数列{an}的各项均为正数,且a2与a4的等比中项为,a2与a3的等差中

等比数列{an}的各项均为正数,且a2与a4的等比中项为,a2与a3的等差中项为

(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.

答案

解:

(Ⅰ)依题意得:a2a4=()2

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

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