Question

1+1/（1+2）+1/（1+2+3）+1/（1+2+3+4）+...1/（1+2+3+...20）怎么算,要写出解题过程

Answer 1

1+2+3+...+n=(1+n)*n/2.
1+1/（1+2）+1/（1+2+3）+1/（1+2+3+4）+...1/（1+2+3+...20）
=1+1/[2(2+1)/2]+1/[3(3+1)/2]+1/[4(4+1)/2]+...+1/[20(20+1)/2]
=1+2/(2*3)+2/(3*4)+2/(4*5)+...+2/(20*21)
=1+2[1/(2*3)+1/(3*4)+1/(4*5)+...+1/(20*21)]
=1+2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/20-1/21)
=1+2(1/2-1/21)
=1+2(19/42)
=1+(19/21)
=40/21.

Answer 2

1/(3+4+……+n)=2/((n+3)*(n-2))=0.4*(1/(n-2)-1/(n+3))
1/3=0.4*(1-1/6)
1/(3+4)=0.4*(1/2-1/5)
类推
1/(3+4+……+20)=0.4*(1/18-1/23)
所以原式
=0.4*((1+1/2+1/3+……+1/18)-(1/6+1/7+……+1/23))
=0.4*(1+1/2+1/3+1/4+1/5-1/19-1/20-1/21-1/22-1/23)

Answer 3

1+2=(1+2)*2/2=3*2/2
1+2+3=(1+3)*3/2=4*3/2
1+2+3...+n=(1+n)n/2
原式=2/(1*2)+2/(2*3)+2/(3*4)+...+2/(20*21)
=2*(1-1/2+1/2-1/3+1/3-1/4......+1/20-1/21)
=2*(1-1/21)
=40/21

Answer 4

1+2+3+...+n=n(n+1)/2
1/（1+2+3+...+n）=2/n(n+1)=2(1/n-1/n+1)
1+1/（1+2）+1/（1+2+3）+1/（1+2+3+4）+...1/（1+2+3+...20）
=2(1-1/2+1/2-1/3+1/3-1/4+...+1/20-1/21)=40/21

Answer 5

An=2/n(n+1)=2(1/n-1/n+1)

则原式=1+2/2-2/3+2/3-2/4+2/4-2/5....+2/20-2/21=19/21