一道分式数学题,急~

显示全部楼层 · 2009-7-9 17:28:48

1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.....+1/(a+2004))(a+2005)=1/a-1/(a+1)+1/(a+1)-1/(a+2)+..-1/(a+2004)+1/(a+2004)-1/(a+2005)=1-1/(a+2005)=(a+2004)/(a+2005)...

千问 · 2009-7-9 17:28:48

给你个公式:1/n(n+1)=1/n-1/(n+1)。所以原式=2005/[a(a+2005)]。...

千问 · 2009-7-9 17:28:48

分数太少了，但是我还是要说明一下方法了，助人为乐嘛！呵呵！用裂项法，原式＝1/a-1/(a+1)+...+1/(a+2004)-1/(a+2005)=1/a-1/(a+2005)...

千问 · 2009-7-9 17:28:48

∵1/a(a+1)=1/a -1/(a+1)1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.....+1/(a+2004))(a+2005)∴原式=1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.....+1/(a+2004))(a+2005)=1/a-1/(a+1)+1/(a+1)-1...