广州大学生校鸡怎么找:简单计算
来源:百度文库 编辑:高考问答 时间:2024/05/04 07:02:16
设S=(1+1/1)-(1/2)+(2+1/2)-(1/3)+(3+1/3)-(1/4)+...+(2004+1/2004)-(1/2005)
求与S最接近的整数
要有过程
(2+1/2)的意思是2+1的和/2
求与S最接近的整数
要有过程
(2+1/2)的意思是2+1的和/2
S=(1+1/1)-(1/2)+(2+1/2)-(1/3)+(3+1/3)-(1/4)+...+(2004+1/2004)-(1/2005)
=1+1/1-1/2+2+1/2-1/3+3+1/3-1/4+...+2004+1/2004-1/2005
=1+1+2+3+4+…+2004-1/2005
=1+(1+2004)*2004/2-1/2005
=2009011-1/2005
≈2009011
首先要去括号,那么(-1/2)和(+1/2)=0、(-1/3)和(+1/3)也=0……所以分数只剩下(-1/2005)
原式=1+(1+2004)*2004除以2-1/2004=2009010又2004/2005≈2009011
化简结果就是 S=1+1+2+3+4+...+2004-1/2005=2009010+1-1/2005=2009010+2004/2005
所以结果是2009011
原式=(1+2004)*2004/2+1-1/2005=2009011-1/2005
所以最接近的当然是2009011
2009010又2004/2005 所以过程是S=(1+1/1)-(1/2)+(2+1/2)-(1/3)+(3+1/3)-(1/4)+...+(2004+1/2004)-(1/2005)
=1+1/1-1/2+2+1/2-1/3+3+1/3-1/4+...+2004+1/2004-1/2005
=1+1+2+3+4+…+2004-1/2005
=1+(1+2004)*2004/2-1/2005
=2009011-1/2005
≈2009011