Question

1+1/(1+2)+1/(1+2+3)+···1/(1+2+3+···+100)等于几？

Answer 1

1+2+3+……+n=n(n+1)/2
1/(1+2+3+……+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
所以1+1/(1+2)+1/(1+2+3)+···1/(1+2+3+···+100)
=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

Answer 2

1/(1+1)*1/2+1/(1+2)*2/2+1/(1+3)*3/2+……+1/(1+n)n/2
=1/2[1/1*2+1/2*3+1/3*4+1/n*(n+1)]
=1/2[1/1-1/2+1/2-1/3+1/3-1/4+……-1/n+1/n-1/(n+1)]
=1/2[1-1/(n+1)]
=1/2*n/(n+1)
………n
==--------
……2(n+1)
这是一般公式，我告诉你解题的技巧，以不变应万变！加油！

Answer 3

1.
1+2+3+……+n=n(n+1)/2
1/(1+2+3+……+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
所以1+1/(1+2)+1/(1+2+3)+···1/(1+2+3+···+100)
=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
2.
通项为1/(1+2+…+n)=1/[(n+1)n/2]=2/[n(n+1)]
=2[1/n-1/(n+1)]
所以
1+1/(1+2)+……+1/(1+2+…+n)
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)];
代入n=100得到
2[1-1/101]=200/101

Answer 4

通项为1/(1+2+…+n)=1/[(n+1)n/2]=2/[n(n+1)]
=2[1/n-1/(n+1)]
所以
1+1/(1+2)+……+1/(1+2+…+n)
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)];
代入n=100得到
2[1-1/101]=200/101