1)1/1×6+1/6×11+1/11×16+1/16×212)1/8+1/24+1/48+1/80+1/1203)(1/2+1/3+1/2004)(1+1/2+.+1/2003)-(1+1/2+.1/2014)(1/2+1/3+.+1/2003)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/14 13:11:25
1)1/1×6+1/6×11+1/11×16+1/16×212)1/8+1/24+1/48+1/80+1/1203)(1/2+1/3+1/2004)(1+1/2+.+1/2003)-(1+1/2+.1/2014)(1/2+1/3+.+1/2003)

1)1/1×6+1/6×11+1/11×16+1/16×212)1/8+1/24+1/48+1/80+1/1203)(1/2+1/3+1/2004)(1+1/2+.+1/2003)-(1+1/2+.1/2014)(1/2+1/3+.+1/2003)
1)1/1×6+1/6×11+1/11×16+1/16×21
2)1/8+1/24+1/48+1/80+1/120
3)(1/2+1/3+1/2004)(1+1/2+.+1/2003)-(1+1/2+.1/2014)(1/2+1/3+.+1/2003)

1)1/1×6+1/6×11+1/11×16+1/16×212)1/8+1/24+1/48+1/80+1/1203)(1/2+1/3+1/2004)(1+1/2+.+1/2003)-(1+1/2+.1/2014)(1/2+1/3+.+1/2003)
1)1/1×6+1/6×11+1/11×16+1/16×21
=1/5(5/1×6+5/6×11+5/11×16+5/16×21)
=1/5×(1-1/6+1/6-1/11+1/11-1/16+1/16-1/21)
=1/5×(1-1/21)
=1/5×20/21
=4/21
2)1/8+1/24+1/48+1/80+1/120
=1/4(1/2+1/6+1/12+1/20+1/50)
=1/4(1/1×2+1/2×3+1/3×4+1/4×5+1/5×6)
=1/4×(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6)
=1/4×(1-1/6)
=5/24
3)(1/2+1/3+……+1/2004)(1+1/2+.+1/2003)-(1+1/2+.1/2014)(1/2+1/3+.+1/2003)
设1/2+1/3+.+1/2003=a
则原代数式可化为
(a+1/2014)(1+a)-(1+a+1/2014)a
=a+a²+1/2014+a/2014-a-a²-a/2014
=1/2014

(1)1/1×6=(1/5)×(1-1/6)
1/6×11=(1/5)×(1/6-1/11)
1/11×16=(1/5)×(1/11-1/16)
1/16×21=(1/5)×(1/16-1/21)
1/1×6+1/6×11+1/11×16+1/16×21
=(1/5)×(1-1/6+1/6-1/11+1/11-1/16+1/16-1/21)=4/21

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(1)1/1×6=(1/5)×(1-1/6)
1/6×11=(1/5)×(1/6-1/11)
1/11×16=(1/5)×(1/11-1/16)
1/16×21=(1/5)×(1/16-1/21)
1/1×6+1/6×11+1/11×16+1/16×21
=(1/5)×(1-1/6+1/6-1/11+1/11-1/16+1/16-1/21)=4/21
(2)1/8+1/24+1/48+1/80+1/120
=1/4×(1/2+1/6+1/12+1/20+1/30)
=1/4×(1/1*2+1/2*3+1/3*4+1/4*5+1/5*6)
=1/4×(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6)
=1/4×(1-1/6)
=1/4×5/6
=5/24

收起

第二小问是三分之一的等比数列