∵Sn=3n+r,Sn-1=3n-1+r,(n≥2,n∈N+),
∴an=Sn-Sn-1=2•3n-1,
又a1=S1=3+r,由通项得:a2=6,公比为3,
∴a1=2,
∴r=-1.
故选B
若等比数列{an}的前n项和Sn=3n+r,则r=( ) A.0 B.-1 C.1 D.3
若等比数列{an}的前n项和Sn=3n+r,则r=( )
A. 0
B. -1
C. 1
D. 3
A. 0
B. -1
C. 1
D. 3
数学人气:601 ℃时间:2019-10-23 11:29:02
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