分析 函數(shù)f(n)=\left\{\begin{array}{l}{n^2}(當(dāng)n為奇數(shù)時)\\-{n^2}(當(dāng)n為偶數(shù)時)\end{array}\right.且an=f(n)+f(n+1),可得a2n=f(2n)+f(2n+1)=4n+1,a2n-1=f(2n-1)+f(2n)=1-4n.可得a2n+a2n-1=2.即可得出.
解答 解:∵函數(shù)f(n)=\left\{\begin{array}{l}{n^2}(當(dāng)n為奇數(shù)時)\\-{n^2}(當(dāng)n為偶數(shù)時)\end{array}\right.且an=f(n)+f(n+1),
∴a2n=f(2n)+f(2n+1)=-(2n)2+(2n+1)2=4n+1,
a2n-1=f(2n-1)+f(2n)=(2n-1)2-(2n)2=1-4n.
∴a2n+a2n-1=2.
則a1+a2+…+a99=(a1+a2)+(a3+a4)+…+(a97+a98)+a99
=2×49+1-4×50=-101.
故選:C.
點評 本題考查了數(shù)列遞推關(guān)系、分組求和,考查了推理能力與計算能力,屬于中檔題.