裂项相消1/(1x3)+1/(3x5)+1/(5x7)+......+1/(2005x2007)=(1-1/3+1/3-1/5+1/5-1/7+......+1/2005-1/2007)/2=(1-1/2007)/2=(2006/2007)/2=1003/2007。
2024-07-04 14:19:14 来源: 编辑:
裂项相消1/(1x3)+1/(3x5)+1/(5x7)+......+1/(2005x2007)=(1-1/3+1/3-1/5+1/5-1/7+......+1/2005-1/2007)/2=(1-1/2007)/2=(2006/2007)/2=1003/2007。