1/1x3+1/3x5+1/5x7+.1/2005x2007怎么算?说明思路.
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1/1x3+1/3x5+1/5x7+.1/2005x2007怎么算?说明思路.
1/1x3+1/3x5+1/5x7+.1/2005x2007怎么算?说明思路.
1/1x3+1/3x5+1/5x7+.1/2005x2007怎么算?说明思路.
1/1x3+1/3x5+1/5x7+.1/2005x2007
=1/2[(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/2005-1/2007)]
=1/2(1/1-1/2007)
=1003/2007
1/(1x3)=(1/1-1/3)/2
1/[nx(n+2)]=[1/n-1/(n+2)]/2
1/1x3+1/3x5+1/5x7+......1/2005x2007
=1/2[(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/2005-1/2007)]
=1/2(1/1-1/2007)
=1003/2007
用裂项相加减法。
将各分式裂项。如,
1/1*3=0.5(1-1/3)
1/3*5=0.5(1/3-1/5)
1/5x7=0.5(1/5-1/7)
……………………
1/2005*2007=0.5(1/2005-1/2007)
各式相加,得
1/1x3+1/3x5+1/5x7+......1/2005x2007
=0.5(1-...
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用裂项相加减法。
将各分式裂项。如,
1/1*3=0.5(1-1/3)
1/3*5=0.5(1/3-1/5)
1/5x7=0.5(1/5-1/7)
……………………
1/2005*2007=0.5(1/2005-1/2007)
各式相加,得
1/1x3+1/3x5+1/5x7+......1/2005x2007
=0.5(1-1/3)+0.5(1/3-1/5)+(1/5-1/7)+……+0.5(1/2005-1/2007),提出0.5,去括号得。
=0.5(1-1/3+1/3-1/5+1/5-1/7+……+1/2005-1/2007)
=0.5(1-1/2007)=1003/2007
(0.5变成1/2容易看一点)
收起
用裂项相加减法。
将各分式裂项。如,
1/1*3=0.5(1-1/3)
1/3*5=0.5(1/3-1/5)
1/5x7=0.5(1/5-1/7)
……………………
1/2005*2007=0.5(1/2005-1/2007)
各式相加,得
1/1x3+1/3x5+1/5x7+......1/2005x2007
全部展开
用裂项相加减法。
将各分式裂项。如,
1/1*3=0.5(1-1/3)
1/3*5=0.5(1/3-1/5)
1/5x7=0.5(1/5-1/7)
……………………
1/2005*2007=0.5(1/2005-1/2007)
各式相加,得
1/1x3+1/3x5+1/5x7+......1/2005x2007
=0.5(1-1/3)+0.5(1/3-1/5)+(1/5-1/7)+……+0.5(1/2005-1/2007),提出0.5,去括号得。
=0.5(1-1/3+1/3-1/5+1/5-1/7+……+1/2005-1/2007)
=0.5(1-1/2007)=1003/2007
收起
用裂项相加减法。
将各分式裂项。如,
1/1*3=0.5(1-1/3)
1/3*5=0.5(1/3-1/5)
1/5x7=0.5(1/5-1/7)
……………………
1/2005*2007=0.5(1/2005-1/2007)
各式相加,得
1/1x3+1/3x5+1/5x7+......1/2005x2007
全部展开
用裂项相加减法。
将各分式裂项。如,
1/1*3=0.5(1-1/3)
1/3*5=0.5(1/3-1/5)
1/5x7=0.5(1/5-1/7)
……………………
1/2005*2007=0.5(1/2005-1/2007)
各式相加,得
1/1x3+1/3x5+1/5x7+......1/2005x2007
=0.5(1-1/3)+0.5(1/3-1/5)+(1/5-1/7)+……+0.5(1/2005-1/2007),提出0.5,去括号得。
=0.5(1-1/3+1/3-1/5+1/5-1/7+……+1/2005-1/2007)
=0.5(1-1/2007)=1003/2007
收起
原式=(1-1/3+1/3-1/5+1/5+……+1/2005-1/2007)÷2
=2006/2007÷2
=1003/2007
用裂项相加减法。
将各分式裂项。如,
1/1*3=0.5(1-1/3)
1/3*5=0.5(1/3-1/5)
1/5x7=0.5(1/5-1/7)
……………………
1/2005*2007=0.5(1/2005-1/2007)
各式相加,得
1/1x3+1/3x5+1/5x7+......1/2005x2007
全部展开
用裂项相加减法。
将各分式裂项。如,
1/1*3=0.5(1-1/3)
1/3*5=0.5(1/3-1/5)
1/5x7=0.5(1/5-1/7)
……………………
1/2005*2007=0.5(1/2005-1/2007)
各式相加,得
1/1x3+1/3x5+1/5x7+......1/2005x2007
=0.5(1-1/3)+0.5(1/3-1/5)+(1/5-1/7)+……+0.5(1/2005-1/2007),提出0.5,去括号得。
=0.5(1-1/3+1/3-1/5+1/5-1/7+……+1/2005-1/2007)
=0.5(1-1/2007)=1003/2007
收起
相邻的奇数互质哇.
两个互质数a,b(a大于b)符合
1/b-1/a=(a-b)/(ab).
解:原式
=0.5(1-1/3)+0.5(1/3-1/5)+0.5(1/5-1/7)+...+0.5(1/2005-1/2007)
=0.5(1-1/2007)(中间已消去)
=1003/2007.