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
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
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.