1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……1/(1+2+3+……+2003)

1+2+……+n=n(n+1)/2
所以1/(1+2+……+n)=2/n(n+1)=2*[1/n-1/(n+1)]

所以1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……1/(1+2+3+……+2003)
=2*(1/1-1/2)+2*(1/2-1/3)+……+2*(1/2003-1/2004)
=2*[(1/1-1/2)+(1/2-1/3)+……+(1/2003-1/2004)]
=2*(1-1/2004)
=2003/1002

______1______ _____2_____ _1_ __1__
An= __(1+n)n__ = (1+n)n =2( n - n+1 )
2
原式=1+2(1/2-1/3+1/3-1/4+…+1/2003-1/2004)
=1+2(1/2-1/2004)
=1+1001/1002
=2003/1002

我只记得计算方法了，很久没有算过了。
用拆项消去法计算