数列{an}满足an+an+1=1/2(n属于N*),a1=-1/2,Sn是{an}的前n项和,则S2011=_
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数列{an}满足an+an+1=1/2(n属于N*),a1=-1/2,Sn是{an}的前n项和,则S2011=_
数列{an}满足an+an+1=1/2(n属于N*),a1=-1/2,Sn是{an}的前n项和,则S
2011=_
数列{an}满足an+an+1=1/2(n属于N*),a1=-1/2,Sn是{an}的前n项和,则S2011=_
a(n) + a(n+1)=1/2 =>任意连续两数是互补成0.5的互补数
a1=-0.5 =>a1的补数 a2=1
a3跟a1同为a2的补数=>a3=a1
如此类推,a(单数)=-0.5
相似地,a(双数)=1
s2011=(a1+...+a2011)+(a2+...+a2010)=1006*(-0.5)+1005*1=502
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