题目内容
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.10200 |
∵f(n)=n2cos(nπ)=
=(-1)n•n2,
由an=f(n)+f(n+1)
=(-1)n•n2+(-1)n+1•(n+1)2
=(-1)n[n2-(n+1)2]
=(-1)n+1•(2n+1),
得a1+a2+a3+…+a100=3+(-5)+7+(-9)+…+199+(-201)=50×(-2)=-100.
故选B
|
由an=f(n)+f(n+1)
=(-1)n•n2+(-1)n+1•(n+1)2
=(-1)n[n2-(n+1)2]
=(-1)n+1•(2n+1),
得a1+a2+a3+…+a100=3+(-5)+7+(-9)+…+199+(-201)=50×(-2)=-100.
故选B
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