由题意:
a1=S1=2+a
a2=S2-S1=(4+a)-(2+a)=2
a3=S3-S2=(8+a)-(4+a)=4
因为 {an}是等比数列
所以 a1=1,即 a=-1
由题意:
b1=T1=b-1
b2=T2-T1=b-(b-1)=1
b3=T3-T2=(3+b)-b=3
因为 {bn}是等差数列
所以 b1=-1,即 b=0
则 a+b=-1
a1=a1005-1004d
a2=a1005-1003d
a3=a1005-1002d
a2007=a1005+1002d
a2008=a1005+1003d
a2009=a1005+1004d
所以 a1+a2+a3+...+a2009=a1005*2009=0
所以 a1005=0
所以 a4+a2006=(a1005-1001d)+(a1005+1001d)=a1005*2=0
同理 a1+a2009=0; a5+a2005=0
选A
设等比数列{an}的前n项和Sn=2∧n+a,等差数列{bn}的前n项和Tn=n²-2n+b,则a+b=_
设等比数列{an}的前n项和Sn=2∧n+a,等差数列{bn}的前n项和Tn=n²-2n+b,则a+b=_
2.已知等差数列{an}满足a1+a2+a3+…a2009=0,则有( )
Aa5+a2005=0 Ba4+a2006
2.已知等差数列{an}满足a1+a2+a3+…a2009=0,则有( )
Aa5+a2005=0 Ba4+a2006
数学人气:102 ℃时间:2020-03-25 14:09:28
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