计算：1/1×3+1/3×5+1/5×7+1/7×9``````+1/2001×2003

计算：1/1×3+1/3×5+1/5×7+1/7×9``````+1/2001×2003

掌握方法,更重要!观察各项得：每项类似1/[a*（a+2）]由以前1/[a*（a+1）]=1/a-1/(a+1)的经验,发现1/a-1/(a+2)=2/[a*（a+2）]∴1/[a*（a+2）]=（1/2）*【1/a-1/(a+2)】然后用1/1*2+1/2*3+.+1/a*(a+1)类似的方法,可得：引用下 我不是他舅 的回答【=(1/2)×(1-1/3)+(1/2)×(1/3-1/5)+(1/2)×(1/5-1/7)+……+(1/2)×(1/2001-1/2003)=(1/2)×(1-1/3+1/3-1/5+1/5-1/7+……+1/2001-1/2003)=(1/2)×(1-1/2003)=1001/2003 】======以下答案可供参考======供参考答案1:=(1/2)×(1-1/3)+(1/2)×(1/3-1/5)+(1/2)×(1/5-1/7)+……+(1/2)×(1/2001-1/2003)=(1/2)×(1-1/3+1/3-1/5+1/5-1/7+……+1/2001-1/2003)=(1/2)×(1-1/2003)=1001/2003供参考答案2:=(1/1-1/3+1/3-1/5......+1/2001-1/2003)/2=1001/2003(1/a-1/(a+2))=(1/a-1/(a+2))/2供参考答案3:1-1/3=2/31/1×3=1/2(1-1/3)1/3-1/5=5/3×5-3/3×5=(5-3)/3×5=2/3×5 通分 相减 都为两数积的两倍1/3×5=1/2(1/3-1/5)1/5-1/7=7/5×7-5/5×7=(7-5)/5×7=2/5×71/5×7=1/2(1/5-1/7)......原式=1/2(1-1/3+1/3-1/5+1/5-1/7+.....-1/2001+1/2001-1/2003) =(1/2)×(1-1/2003) =1001/2003

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