(1)1/2*3+1/3*4+1/4*5.1/2006*2005(2)1/2+1/(1+1)*(2+1)+1/(1+2)*(2+2).1/(1+2009)*(2+2009)(3)1+1/(1+2)+1/(1+2+3)+.1/(1+2+3+4.100)

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/28 10:29:34

(1)1/2*3+1/3*4+1/4*5.1/2006*2005(2)1/2+1/(1+1)*(2+1)+1/(1+2)*(2+2).1/(1+2009)*(2+2009)(3)1+1/(1+2)+1/(1+2+3)+.1/(1+2+3+4.100)
(1)1/2*3+1/3*4+1/4*5.1/2006*2005
(2)1/2+1/(1+1)*(2+1)+1/(1+2)*(2+2).1/(1+2009)*(2+2009)
(3)1+1/(1+2)+1/(1+2+3)+.1/(1+2+3+4.100)

(1)1/2*3+1/3*4+1/4*5.1/2006*2005(2)1/2+1/(1+1)*(2+1)+1/(1+2)*(2+2).1/(1+2009)*(2+2009)(3)1+1/(1+2)+1/(1+2+3)+.1/(1+2+3+4.100)
(1)1/2*3+1/3*4+1/4*5.1/2006*2005
=1/2 -1/3+1/3-1/4+1/4-1/5+.+/12005-1/2006
=1/2 - 1/2006
=1002/2006 自己化简一下.
(2)1/2+1/(1+1)*(2+1)+1/(1+2)*(2+2).1/(1+2009)*(2+2009)
=1/2+1/2-1/3+1/3-1/4+.+1/12010-1/2011
=1- 1/2011
=2010/2011
3)1+1/(1+2)+1/(1+2+3)+.1/(1+2+3+4.100)
=1+ 2 *(1/2-1/3)+2*(1/3-1/4)+.+2*(1/100-1/101)
=1+2*(1/2-1/3+1/3-1/4+.+1/100-1/101)
=1+2*(1/2-1/101)
=2-2/101
=200/101

1)原式=1/2-1/3+1/3-1/4+...+1/2005-1/2006=1/2-1/2006=501/1003
2)原式=1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011=1-1/2011=2010/2011
3) 可以把分母看成等差数列的前n项和=n(n+1)/2
而2/【n(n+1)】=2[1/n-1/(n+1)]
所以原式=2{1-1/2+1/2-1/3+...+1/n-1/(n+1)}=2[1-1/(n+1)]=2n/(n+1)
令n=100即可

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