已知数列{an}的前n项和Sn=2an-2^n

显示全部楼层 · 2016-12-2 03:08:54

(1)求a3,a4
(2)证明数列{a(n+1)-2an}是等比数列.
(3)求{an}的通项公式
十万火急
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千问 · 2016-12-2 03:08:54

解;(1)s1=a1=2a1-2a1=2s2=a1+a2=2a2-4a2=6s3=a1+a2+a3=2a3-8a3=16s4=a1+a2+a3+a4=2a4-16a4=40(2)Sn+1=2an+1-2^(n+1)Sn=2an-2^n相减,an+1=2an+1-2an-2^na(n+1)-2an=2^n设bn=a(n+1)-2an则有:bn+1/bn=2(常数)b1=a2-2a1=2所以{bn}是以2为首项2为公比的等比数列数列{a(n+1)-2an}是等比数列. (3)a(n+1)-2an=2^n,an-2an-1=2^(n-1),->2a...

千问 · 2016-12-2 03:08:54

Sn=2an-2^nSn-1=2an-1-2^(n-1)Sn-Sn-1=an=2an-2an-1-2^(n-1)an=2an-1+2^(n-1)a1=2a1-2a1=2a2=2a1+2=6a3=16a4=40an=2an-1+2^(n-1)an-2an-1=2^(n-1)a(n+1)-2an=2^n(a...

千问 · 2016-12-2 03:08:54

1、a1=2a1-2
=>
a1=2a2=S1-a1=2a2-4-2 => a2=6a3=S3-S2=2a3-8-8 => a3=16a4=S4-S3=2a4-16-24 => a4=402、a(n+1)=S(n+1)-Sn 2an=Sn-2^n a(n+1)-2an=S(n+1)-Sn-S...