1/(2×4)+1/(4×6)+1/(6×8)+1/(8×10)+·····+1/(2008×2010)的算法(详细过程),

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1/(2×4)+1/(4×6)+1/(6×8)+1/(8×10)+·····+1/(2008×2010)的算法(详细过程),

1/(2×4)+1/(4×6)+1/(6×8)+1/(8×10)+·····+1/(2008×2010)的算法(详细过程),
1/(2×4)+1/(4×6)+1/(6×8)+1/(8×10)+·····+1/(2008×2010)的算法(详细过程),

1/(2×4)+1/(4×6)+1/(6×8)+1/(8×10)+·····+1/(2008×2010)的算法(详细过程),
原式=1/4{4×1/(2×4)+4×1/(4×6)+……4×1/2008×2010}
=1/4{1/1×2+1/2×3+……1/1004×1005}
=1/4×(1004/1005)
=251/1005
解释一下
4×1/(2×4)+4×1/(4×6)+……4×1/2008×2010把4分成2×2分别和分母的两个偶数约分
{1/1×2+1/2×3+……1/1004×1005}这是都很熟悉的了得n/n+1

原式=2x(1/2-1/4)+2x(1/4-1/6)+2x(1/6-1/8)+2x(1/8-1/10)+......+2x(1/2008-1/2009)=2x(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+......+1/2008-1/2009)=2x(1/2-1/2009)=2007/2009

把1/2提出来,就剩1-1/2+1/2-1/3+1/3....-1/1005,答案就是502/1005

原式=1/2*(1/2-1/4)+1/2*(1/4-1/6)+……+1/2*(1/2008-1/2010)
=1/2*(1/2-1/4+1/4-1/6+1/6-1/8+……+1/2008-1/2010)
=1/2*(1/2-1/2010)
=1/2*1004/2010
=502/2010
=251/1005