别克英朗code82复位:1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3...+n)=?

来源：百度文库 编辑：中科新闻网 时间：2024/05/02 13:49:32

1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3...+n)
=1+1*2/(2*3)+1*2/(3*4)+...+1*2/[n*(1+n)]
=2[1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/(n+1)]
=2[1/2+1/2-1/(n+1)]
=2-2/(n+1)
=2n/(n+1)

(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) 1+1/1+2/=?? 1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100) 1+1/1+2+1/1+2+3+......+1/1+2+3+......+10 1/1+1/2+1/3+1/4+。。。。+1/N 是多少 (1-1/2^2)(1-1/3^2)....(1-1/10^2) (1-1/2^2)(1-1/3^2)....(1-1/10^2) s=1/1+1/1!+1/2！+1/3！+1/4！+....+1/25! (1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)*……*（1-1/99）*（1+1/99）。 (1+1/2)*(1-1/2)(1+1/3)*(1-1/3)*.......(1+1/99))*(1-1/99)怎么做