题目内容
已知函数且an=f(n)+f(n+1),则a1+a2+a3+…+a100等于( )
A.0 |
B.100 |
C.-100 |
D.10200 |
B
由题意,a1+a2+a3+…+a100
=12-22-22+32+32-42-42+52+…+992-1002-1002+1012
=-(1+2)+(3+2)+…-(99+100)+(101+100)
=-(1+2+…+99+100)+(2+3+…+100+101)
=-1+101
=100,选B.
=12-22-22+32+32-42-42+52+…+992-1002-1002+1012
=-(1+2)+(3+2)+…-(99+100)+(101+100)
=-(1+2+…+99+100)+(2+3+…+100+101)
=-1+101
=100,选B.
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