题目

已知函数f(n)=n2cos(nπ),且an=f(n)+f(n+1),则a1+a2+a3+…+a100=( ) A.0                 B.-100 C.100                   D.10 200 答案：B.因为f(n)=n2cos(nπ), 所以a1+a2+a3+…+a100=[f(1)+f(2)+…+f(100)]+[f(2)+…+f(101)], f(1)+f(2)+…+f(100) =-12+22-32+42-…-992+1002 =(22-12)+(42-32)+…+(1002-992) =3+7+…+199 ==5 050, f(2)+…+f(101) =22-32+42-…-992+1002-1012 =(22-32)+(42-52)+…+(1002-1012) =-5-9-…-201==-5 150, 所以a1+a2+a3+…+a100 =[f(1)+f(2)+…+f(100)]+[f(2)+…+f(101)] =-100.判断函数y=-x3+1在R上的单调性并给予证明．