求着到奥林匹克数学题的答案以及解题方式.
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/06/28 06:50:53
求着到奥林匹克数学题的答案以及解题方式.
![](http://img.wesiedu.com/upload/7/11/711f96ac6fb86e3cee86e9664a36c8f8.jpg)
就是那个最后一题.
![](http://img.wesiedu.com/upload/7/11/711f96ac6fb86e3cee86e9664a36c8f8.jpg)
就是那个最后一题.
![求着到奥林匹克数学题的答案以及解题方式.](/uploads/image/z/2612544-24-4.jpg?t=%E6%B1%82%E7%9D%80%E5%88%B0%E5%A5%A5%E6%9E%97%E5%8C%B9%E5%85%8B%E6%95%B0%E5%AD%A6%E9%A2%98%E7%9A%84%E7%AD%94%E6%A1%88%E4%BB%A5%E5%8F%8A%E8%A7%A3%E9%A2%98%E6%96%B9%E5%BC%8F.)
法一:
设:A=2010/1-2009/2+2008/3-2007/4+2006/5-2005/6+.+2/2009-1/2010
B=1/1006+3/1007+5/1008+.+2009/2010
2010/1-2009/2=2011/(1×2)
2008/3-2007/4=2011/(3×4)
2006/5-2005/6=2011/(5×6)
.
2/2009-1/2010=2011/(2009×2010)
A=2011×[1/(1×2)+1/(3×4)+...+1/(2009×2010)]
=2011×(1-1/2+1/3-1/4+1/5-...+1/2009-1/2010)
=2011*(1+1/2+1/3+1/4+1/5+1/6+……+1/2009+1/2010)-2*(1/2+1/4+1/6+.+1/2010)
=2011*(1+1/2+1/3+1/4+1/5+1/6+.+1/1005)+(1/1006……+1/2009+1/2010)-(1+1/2+1/3+.+1/1005)
=2011*(1/1006+1/1007+1/1008+.+1/2010)
所以:
A+B=(2011+1)/1006+(2011+3)/1007+...+(2011+2009)/2010
=2012/1006+2014/1007+.+4020/2010
=2+2+2.+2(一共有1005个2相加)
=2×1005
法二:
2010/1-2009/2+2008/3-2007/4+...-1/2010+1/1006+3/1007+5/1008+...+2009/2010
=(2010/1+1)-(2009/2+1)+...-(1/2010+1)+(1/1006-2)+(3/1007-2)+...+(2009/2010-2)+2010
=2011/1-2011/2+...-2011/2010-(2011/1006+2011/1007+...+2011/2010)+2010
=2011*(1-1/2+1/3-1/4+...+1/2009-1/2010)-2011*(1/1006+1/1007+...+1/2010)+2010
=2011*(1+1/2+1/3+...+1/2010-2*(1/2+1/4+...+1/2010)-1/1006-1/1007-...-1/2010)+2010
=2011*(1+1/2+1/3+...+1/2010-1-1/2-1/3-1/1005-1/1006-1/1007-...-1/2010)+2010
=0+2010
=2010
设:A=2010/1-2009/2+2008/3-2007/4+2006/5-2005/6+.+2/2009-1/2010
B=1/1006+3/1007+5/1008+.+2009/2010
2010/1-2009/2=2011/(1×2)
2008/3-2007/4=2011/(3×4)
2006/5-2005/6=2011/(5×6)
.
2/2009-1/2010=2011/(2009×2010)
A=2011×[1/(1×2)+1/(3×4)+...+1/(2009×2010)]
=2011×(1-1/2+1/3-1/4+1/5-...+1/2009-1/2010)
=2011*(1+1/2+1/3+1/4+1/5+1/6+……+1/2009+1/2010)-2*(1/2+1/4+1/6+.+1/2010)
=2011*(1+1/2+1/3+1/4+1/5+1/6+.+1/1005)+(1/1006……+1/2009+1/2010)-(1+1/2+1/3+.+1/1005)
=2011*(1/1006+1/1007+1/1008+.+1/2010)
所以:
A+B=(2011+1)/1006+(2011+3)/1007+...+(2011+2009)/2010
=2012/1006+2014/1007+.+4020/2010
=2+2+2.+2(一共有1005个2相加)
=2×1005
法二:
2010/1-2009/2+2008/3-2007/4+...-1/2010+1/1006+3/1007+5/1008+...+2009/2010
=(2010/1+1)-(2009/2+1)+...-(1/2010+1)+(1/1006-2)+(3/1007-2)+...+(2009/2010-2)+2010
=2011/1-2011/2+...-2011/2010-(2011/1006+2011/1007+...+2011/2010)+2010
=2011*(1-1/2+1/3-1/4+...+1/2009-1/2010)-2011*(1/1006+1/1007+...+1/2010)+2010
=2011*(1+1/2+1/3+...+1/2010-2*(1/2+1/4+...+1/2010)-1/1006-1/1007-...-1/2010)+2010
=2011*(1+1/2+1/3+...+1/2010-1-1/2-1/3-1/1005-1/1006-1/1007-...-1/2010)+2010
=0+2010
=2010