(1/2+1/3+··1/2010)(1+1/2+1/3··+1/2009)-(1/2+1/3··+1/2009)(1+
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/19 09:59:45
(1/2+1/3+··1/2010)(1+1/2+1/3··+1/2009)-(1/2+1/3··+1/2009)(1+1/2+1/3··1/2010)等于多少
第一个括号里先+1再-1,也就是变成(1+1/2+1/3+1/4+...+1/2010-1)
式子就变成【(1+1/2+1/3+1/4+...+1/2010)-1】(1+1/2+1/3+...+1/2009)-(1+1/2+1/3+...+1/2010)(1/2+1/3+1/4+...+1/2009)
把前两个乘积按分配律展开,就变成:(1+1/2+1/3+1/4+...+1/2010)(1+1/2+1/3+...+1/2009)-1×(1+1/2+1/3+...+1/2009)-(1+1/2+1/3+...+1/2010)(1/2+1/3+1/4+...+1/2009)
=(1+1/2+1/3+1/4+...+1/2010)(1+1/2+1/3+...+1/2009)-(1+1/2+1/3+...+1/2010)(1/2+1/3+1/4+...+1/2009)-1×(1+1/2+1/3+...+1/2009)
=(1+1/2+1/3+1/4+...+1/2010)【(1+1/2+1/3+...+1/2009)-(1/2+1/3+1/4+...+1/2009)】-1×(1+1/2+1/3+...+1/2009)
=(1+1/2+1/3+1/4+...+1/2010)-(1+1/2+1/3+...+1/2009)
=1/2010
式子就变成【(1+1/2+1/3+1/4+...+1/2010)-1】(1+1/2+1/3+...+1/2009)-(1+1/2+1/3+...+1/2010)(1/2+1/3+1/4+...+1/2009)
把前两个乘积按分配律展开,就变成:(1+1/2+1/3+1/4+...+1/2010)(1+1/2+1/3+...+1/2009)-1×(1+1/2+1/3+...+1/2009)-(1+1/2+1/3+...+1/2010)(1/2+1/3+1/4+...+1/2009)
=(1+1/2+1/3+1/4+...+1/2010)(1+1/2+1/3+...+1/2009)-(1+1/2+1/3+...+1/2010)(1/2+1/3+1/4+...+1/2009)-1×(1+1/2+1/3+...+1/2009)
=(1+1/2+1/3+1/4+...+1/2010)【(1+1/2+1/3+...+1/2009)-(1/2+1/3+1/4+...+1/2009)】-1×(1+1/2+1/3+...+1/2009)
=(1+1/2+1/3+1/4+...+1/2010)-(1+1/2+1/3+...+1/2009)
=1/2010
计算 1*2/1+2*3/1+3*4/1+············+2009*2010/1
1、(1+3+5+···+2009+2011)-(2+4+6+···+2010+2012)
(1)2010×(1-1/2)×(1-1/3)×(1-1/4)×·········×(1-1/2009)×(1-1/20
(-1)+2+(-3)+4+······+(-2009)+2010+(-2011)=__________________
1*2*3···2008*2009*(-1)*(-1/2)*(-1/3)···(-1/2008)*(1/2009)
求证;1+1/2+1/3+······+1/k≧(2k)/(k+1)
2010·(1-1/2)·(1-1/3)·(1-1/4)·…·(1-1/2010)
1/2009×2/2009×3/2009×4/2009×5/2009×······×2008/2009怎么算?
计算: 1/1*2+1/2*3+1/3*4············1/2009*2010 注:* 号代表乘号!
(1+1/2)(1-1/3)(1+1/4)(1-1/5)··········(1-1/2005)(1+1/2006)=_
(1+3+5······+2007+2009+2011)-(2+4+6+·····+2006+2008+2010)
1+2++3+··········+10000=