怎样计算1+1⼀(1+2)+1⼀(1+2+3)+.....+1⼀(1+2+3+....+1999)

1/(1+2+3+....+n)=1/{n(1+n)/2}=2/n(1+n)=2[1/n-1/(1+n)]
1+1/(1+2)+1/(1+2+3)+.....+1/(1+2+3+....+1999)
=1+2(1/2-1/3)+2(1/3-1/4)+……+2（1/1999-1/2000）
=1+2(1/2-1/3+1/3-1/4+1/4……+1/1998-1/1999+1/1999-1/2000)
=1+2(1/2-1/2000)
=1+1-1/1000
=1999/1000

1+2+3+...+n=n(n+1)/2

1+1/(1+2)+1/(1+2+3)+.....+1/(1+2+3+....+1999)
=1+1/(2*3/2)+1/(3*4/2)+.....+1/(1999*2000/2)
=1+2/2*3+2/3*4+.....+2/1999*2000
=1+2(1/2*3+1/3*4+....+1/1999*2000)
=1+2(1-1/2000)
=1+2*1999/2000
=1+1999/1000
=2999/1000

一般项可以写作1/(1+2+....+n)=2/(n(n+1))=2(1/n-1/(n+1))
所以原式=2(1/1-1/2)+2(1/2-1/3)+2(1/3-1/4)+....+2(1/1999-1/2000)
=2(1/1-1/2+1/2-1/3+1/3-1/4+....+1/1999-1/2000)
=2(1-1/2000)
=1999/1000