1/(32-1)+1/(52-1)+1/(72-1)+…+1/[(2n2+1)-1]+…≈( )

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问题 1/(32-1)+1/(52-1)+1/(72-1)+…+1/[(2n2+1)-1]+…≈(    )

选项 A、1/4
B、1
C、1/2
D、无法计算

答案A

解析 (2n+1)2-1=4n2+4n=2n(2n+2),则原式=1/(2×4)+1/(4×6)+1/(6×8)+…+1/[2n×(2n+2)]=1/2(1/2-1/4+1/4-1/6+1/6-1/8+…1/2n-1/(2n+2))=1/2[1/2-1/(2n+2)]≈1/2×1/2=1/4
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