1/2+1/(23)+1/(34)+.+1/(20042005)+1/(20052006)

来源：学生作业帮助网编辑：作业帮时间：2024/05/07 11:52:41

1/2+1/(2*3)+1/(3*4)+.+1/(2004*2005)+1/(2005*2006)
1/2+1/(2*3)+1/(3*4)+.+1/(2004*2005)+1/(2005*2006)

1/2+1/(2*3)+1/(3*4)+.+1/(2004*2005)+1/(2005*2006)
裂项相消法
根据 1/n*(n+1)= 1/n -1/(n+1)
1/2+1/(2*3)+1/(3*4)+.+1/(2004*2005)+1/(2005*2006)
=1/2 +1/2-1/3+1/3-1/4+...+1/2005-1/2006
=2005/2006

1/n(n+1)= 1/n - 1/(n+1)
所以原式=1/2+1/2-1/3+1/3-1/4+1/4-1/5…………-1/2005+1/2005-1/2006
=1-1/2006
=2005/2006

裂项相消法估计是唯一解!

的确是只能用裂项相消法来解了

1+2+3+4+1 1 2 3 4 1 2 3 4 1 2 3 4 1,2,3,4, 1 2 3 4 1 2 3 4 1,2,3,4, 1,2,3,4, 1,2,3,4, 1,2,3,4, 1 2 3 4 1,2,3,4, 1,2,3,4, 1,2,3,4, 1、2、3、4 1+2+3×4 1,2,3,4