设数列{an}前n项和为Sn,且an+Sn=1(n属于N+)

题目

设数列{an}前n项和为Sn,且an+Sn=1(n属于N+)
1.求{an}的同项公式
2.若数列{bn}满足b1=1且b(n+1)=bn+an,求数列{bn}的通项公式

答案

{Sn} = 1 - a|n ①{Sn-1} = 1 - a|n-1 ②① - ② ,可得：a|n = a|n-1 - a|n即a|n = 1/2 a|n-1数列{an}为公比q为1/2的等比数列令n=1则2 a1 = 1,a1 = 1/2根据等比数列公式：通项{an}=[1/2(1- 1/2)]/[1-(1/2)^n](自己化...