设Pn=(1-1/2)(1-1/3)(1-1/4)...[1-1/(n 1)],则1/P2005=?怎么做?

显示全部楼层 · 2010-12-19 23:58:46

千问 · 2010-12-19 23:58:46

Pn=(1-1/2)(1-1/3)(1-1/4)...[1-1/(n 1)],
=[(2-1)/2]*[(3-1)/3]*[(4-1)/4]*.....[(n+1-1)/(n+1)]
=(1/2)*(2/3)*(3/4)*......*[n/(n+1)]
=1/(n+1)1/Pn=n+11/P2005=2006

千问 · 2010-12-19 23:58:46

P2005=(1-1/2)(1-1/3)(1-1/4)...[1-1/2006]=1/2*2/3*3/4....2005/2006=1/20061/P2005=2006

千问 · 2010-12-19 23:58:46

请将括号内的数值用分数算出,左右相消,你就知道怎麼算了