Question

计算题拜托了

1+1/1+2+1/1+2+3+1/1+2+3+4+...+1/1+2+3+4+....+2010=
要写过程啊

Answer 1

1+1/1+2+1/1+2+3+1/1+2+3+4+...+1/1+2+3+4+....+2010=
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+....+2010)=

1+2+3+..+n=n(n+1)/2
所以[1/（1+2+3+…+n)]=2/n(n+1)=2[1/n-1/(n+1)]
所以Sn=1+[1/(1+2)]+〔1/(1+2+3)〕+[1/(1+2+3+4)]+……+[1/(1+2+3+……+n)]
=2[1/1-1/2]+2[1/2-1/3]+2[1/3-1/4]+...+2[1/n-1/(n+1)]
=2[1-1/2+1/2-1/3+1/3-1/4+...+1/(n-1)-1/n+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)

n=2010
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+....+2010)=4020/2011