已知数列an中,a1=2,a(n+1)=1-an,设sn为数列an的前n项和,则s2006-2s2007+s2008=

已知数列an中,a1=2,a(n+1)=1-an,设sn为数列an的前n项和,则s2006-2s2007+s2008=

a(n+1)=1-ana(n+1)-1/2=-(an-1/2)an-1/2=(a1-1/2)(-1)^(n-1)=(3/2)(-1)^(n-1)an=1/2+(3/2)(-1)^(n-1)Sn=[1/2+(3/2)(-1)^0]+[1/2+(3/2)(-1)^1]+[1/2+(3/2)(-1)^2]+……+[1/2+(3/2)(-1)^(n-3)]+[1/2+(3/2)(-1)^(n-2)]+[1/2+(3/2)(-1)^(n-1)]=n/2+(3/2)[(-1)^0+(-1)^1+(-1)^2+……+(-1)^(n-3)+(-1)^(n-2)+(-1)^(n-1)]=n/2+(3/2)[1-(-1)^n]/2Sn=n/2+(3/2)[1-(-1)^n]/2S2006=2006/2+(3/2)[1-(-1)^2006]/2=1003S2007=2007/2+(3/2)[1-(-1)^2007]/2=2007/2+3/2=1005S2008=2008/2+(3/2)[1-(-1)^2008]/2=1004s2006-2s2007+s2008=1003-2010+1004=-3