3.5x5分之4 0.8x7.5-百分之80怎么算
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/31 11:59:25
原式=4/3+16/15+36/35+...+400/399=1/3+1/15+1/35+...+1/399+10=1/3+【(1/3-1/5)+(1/5-1/7)+(1/7-1/9)+...+(1/
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这是sin(x)的Taylor展开,对哪个编程?再问:wintc吧简单点尽量用数组谢谢
1x2分之1-2x3分之1-3x4分之1-...-9x10分之1=1/1-1/2+1/2-1/3+1/3-1/4+1/4-.-1/9+1/9-1/10=1-1/10=9/101x3分之2+3x5分之2
首先把每一个分式拆成两项之差,即1/1×3+1/3×5+1/5×7+……1/99×101=(1/2)×(1-1/3)+(1/2)×(1/3-1/5)+(1/2)×(1/5-1/7)+……+(1/2)×
等号免啦!我来提醒你:1/(2n-1)(2n+1分解为1/(2n-1)-1/(2n+1)所以结果是:1-1/21=20/21
1/1x3+1/3x5+1/5x7+……+1/17x19+1/19x21=1/2*(1-1/3+1/3-1/5+.+1/17-1/19+1/19-1/21)=1/2*(1-1/21)=10/21
=(1-1/3+1/3-1/5+1/5……+1/2011-1/2013)=(1-2013)除以2/2013=2012/2013
1/1x3+1/3x5+1/5x7.1/2009x2011+1/2011x2013=1/2×﹙1-1/3+1/3-1/5······+1/2009-1/2011+1/2011-1/2013)=1/2×
1/(1x3)+1/(3x5)+1/(5x7)+.+1/(2011x2013)=1/2x[2/(1x3)+2/(3x5)+2/(5x7)+.+2/(2011x2013)]=1/2x[1-1/3+1/3
1/1x3+1/3x5+1/5x7+.+1/2009x2011=(1/2)*(1-1/3+1/3-1/5+1/5-1/7_.+1/2009-1/2011)=(1/2)*(1-1/2011)=1005/
=1/2×(1-1/3+1/3-1/5+1/5-1/7+17-1/9+1/9-1/11)=1/2×(1-1/11)=1/2×10/11=5/11
1/1x3+1/3x5+1/5x7+...1/2009x2011+1/2011×2013=(1-1/3+1/3-1/5+1/5-1/7+...+1/2011-1/2013)÷2=(1-1/2013)÷
1/1x3+1/3x5+1/5x7+.+1/1997x1999=(1/2)*(1-1/3+1/3-1/5+1/5-1/7+.+1/1997-1/1999)=(1/2)*(1-1/1999)=999/1
2/1x3+2/3X5+2/5x7+2/7x9+2/9x11=1-1/3+1/3-1/5.+1/9-1/11=1-1/11=10/11
1X3分之1+3X5分之1+5X7分之1+...+2013X2015分之11/1×3=1/2×(1-1/3)1/3×5=1/2×(1/3-1/5).∴原式=1/2×(1-1/3+1/3-1/5.+1/
(71+1/6)×6/7+(61+1/5)×5/6+(51+1/4)×4/5=(70+7/6)×6/7+(60+6/5)×5/6+(50+5/4)×4/5=61+51+41=153如果你觉得我的回答比
1X3分之一+3X5分之一+5X7分之一…+2011X2013分之一=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)...+1/2(1/2011-1/2013)=12/(1-
1分之1x3+1分之3x5+1分之5x7+.+1分之2011x2013=1/(1*3)+1/(3*5)+1/(5*7)+……+1/(2011*2013)=1/2*(1-1/3+1/3-1/5+1/5-
=1/2×(1-1/2011)=1/2×2010/2011=1005/2011再问:为什么这样做呢?