1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+99+100)等于几
问题描述:
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+99+100)等于几
答
1+……+n=n(n+1)/2,所以1/(1+……+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+...+99+100)=2[1/1-1/2+1/2-1/3+……+1/100-1/101]=2(1-1/101)=200/101