已知函数f(n)=n^2（当n为奇数时）或-n^2（当n为偶数时）且an=f(n)+f(n+1),则

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当n=2k-1时,f(n)=n^2即f(2k-1)=(2k-1)^2
当n=2k时,f(n)=-n^2即f(2k)=-(2k)^2
an=f(n)+f(n+1)
a(2k-1)=f(2k-1)+f(2k)
=(2k-1)^2-(2k)^2
=-4k+1
a(2k)=f(2k)+f(2k+1)
=-(2k)^2+(2k+1)^2
=4k+1a(2k-1)+a(2k)=2
S2012=(a1+a2)+(a3+a4)+……+[a(2k-3)+a(2k-2)]+[a(2k-1)+a(2k)]
=2*1006
=2012