(1-2²/1)×(1-3²/1)×...(1-2004²/1)×(1-2005²
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/08 01:04:01
(1-2²/1)×(1-3²/1)×...(1-2004²/1)×(1-2005²/1)
![(1-2²/1)×(1-3²/1)×...(1-2004²/1)×(1-2005²](/uploads/image/z/8600913-9-3.jpg?t=%281-2%26%23178%3B%2F1%29%C3%97%281-3%26%23178%3B%2F1%29%C3%97...%281-2004%26%23178%3B%2F1%29%C3%97%281-2005%26%23178)
平方差
(1-2²/1)×(1-3²/1)×...(1-2004²/1)×(1-2005²/1)
=(1-2/1)×(1+2/1)×(1-3/1)×(1+3/1)...(1-2005/1)×(1+2005/1)
=2/1×2/3×3/2×3/4……2005/2004×2005/2006
中间约分
=2/1×2005/2006
=2005/1003
(1-2²/1)×(1-3²/1)×...(1-2004²/1)×(1-2005²/1)
=(1-2/1)×(1+2/1)×(1-3/1)×(1+3/1)...(1-2005/1)×(1+2005/1)
=2/1×2/3×3/2×3/4……2005/2004×2005/2006
中间约分
=2/1×2005/2006
=2005/1003
(1-2²/1)×(1-3²/1)×...(1-2004²/1)×(1-2005²
计算:1²-2²+3-4²+.+2003²-2004²+2005
1²-2²+3²-4²+...+2003²-2004²+20
1²-2²+3²-4²……+2005²-2006²+2007
1²+2²+3²+4²+5²
计算:(1-1/2²)(1-1/3²)......(1-1/2005²)(1-1/2006
(2³-1²)+(4²-3²)+.(2006²-2005²)
1²-2²+3²-4²+5²-6²+7²-8
2008²-2007²+2006²-2005²…+2²-1²
计算:1²-2²+3²-4²+5²-6²+...+49
1²+2²+3²+.n²等于多少
1²+2²+3²+...+n²的公式,