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

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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
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 】

=(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

=(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

1-1/3=2/3
1/1×3=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=2/5×7
1/5×7=1/2(1/5-1/7)
.....

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1-1/3=2/3
1/1×3=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=2/5×7
1/5×7=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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