设等比数列{an}的前n项和为Sn,已知a1=2011,且an+2an+1+an+2=0,(n∈N*)则S2012=( )A 2011B 2012C 1D 0
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![设等比数列{an}的前n项和为Sn,已知a1=2011,且an+2an+1+an+2=0,(n∈N*)则S2012=( )A 2011B 2012C 1D 0](/uploads/image/z/1741058-26-8.jpg?t=%E8%AE%BE%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5a1%3D2011%2C%E4%B8%94an%2B2an%2B1%2Ban%2B2%3D0%2C%28n%E2%88%88N%2A%29%E5%88%99S2012%3D%28+%29A+2011B+2012C+1D+0)
设等比数列{an}的前n项和为Sn,已知a1=2011,且an+2an+1+an+2=0,(n∈N*)则S2012=( )A 2011B 2012C 1D 0
设等比数列{an}的前n项和为Sn,已知a1=2011,且an+2an+1+an+2=0,(n∈N*)则S2012=( )
A 2011
B 2012
C 1
D 0
设等比数列{an}的前n项和为Sn,已知a1=2011,且an+2an+1+an+2=0,(n∈N*)则S2012=( )A 2011B 2012C 1D 0
an+2a(n+1)+a(n+2)=0
a(n+2)+a(n+1)=-[an+a(n+1)]
a2013+a2012=-(a2012+a2011)=a2011+a2010=...=-(a2+a1)
a2=(a3-a1)/2
a3=(a4-a2)/2
a4=(a5-a3)/2
…………
a2012=(a2013-a2011)/2
累加
a2+a3+a4+...+a2012=[(a3+a4+...+a2013)-(a1+a2+...+a2011)]/2=(a2012+a2013-a1-a2)/2=0
Sn=a1+a2+a3+...+a2012=a1+0=2011+0=2011
选A.