=1/4×[1/（1×3）]（因为越来越接近0，所以我没有计算）
=1/4×1/3
=1/12（注：分母4是（5-1），（7-3）……）
则1/（n+2）*n*（n+4=1/4*｛1/n*(n+2)-1/(n+4)*(n+2)｝
那两项可以消
剩下的自己做吧，做的太多自己的训练效果不好啦！
则1/（n+2）*n*（n+4=1/4*｛1/n*(n+2)-1/(n+4)*(n+2)｝
那两项可以消
=15+175/15+189/15
=15+364/15
=589/15
简便计算1\/1*3*5+1\/3*5*7+1\/5*7*9+1\/7*9*11+1\/9*11*13+1\/11*13*15...
您好！分析:1\/1×3×5=1\/4×(1\/1×3 -1\/3×5)1\/3×5×7=1\/4×(1\/3×5 -1\/5×7)1\/5×7×9=1\/4×(1\/5×7 - 1\/7×9)1\/7×9×11=1\/4×(1\/7×9 -1\/9×11)...1\/11×13×15=1\/4×(1\/11×13 -1\/13×15)所有的等式相加有 1\/1×3×5+1\/3×5×7+1\/...
计算1\/1*3*5+1\/3*5*7+1\/5*7*9+.+1\/2003*2005*2007=多少
1\/1*3*5=(1\/1*3-1\/3*5)\/41\/3*5*7=(1\/3*5-1\/5*7)\/41\/5*7*9=(1\/5*7-1\/7*9)\/4...1\/2003*2005*2007=(1\/2003*2005-1\/2005*2007)\/4全加起来你那个东西就等于(1\/1*3-1\/2005*2007)\/4=335336\/4024035
1\/1×3×5+1\/3×5×7+1\/5×7×9×+……+1\/2001×2003×2005
则原式=(1\\4)×(1\\(1×3)-1\\(3×5)+1\\(3×5)-1\\(5×7)+1\\(7×9)-……+1\\(2001×2003)-1\\(2003×2005))=(1\\4)×(1\\(1×3)-1\\(2003×2005))=(1\\4)×((4016015-3)\\12048045)=1004003\\12048045(结果你自己再算算吧，方法肯定对)...
1\/1×3×5+1\/3×5×7+1\/5×7×9+···+1\/2001×2003×2005等于多少...
=1\/12-1\/(4×2003×2005)。公式：(∑是求和的记号）∑1\/(2n-1)(2n+1)(2n+3)=1\/4 - 1\/[4(2n+1)(2n+3)] .
1\/1×3+1\/3×5+1\/5×7+1\/7×9+1\/9×11 = 多少?? 过程也写详细一点不要...
因为1\/1×3=1\/1-1\/3，所以以此类推，1\/(n-2)-1\/n=1\/(nx(n-2))，所以可得 1\/1×3＋1\/3×5+1\/5×7+1\/7×9+1\/9×11 =1\/1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11=1\/1-1\/11=10\/11
1\/1*3+1\/3*5+1\/5*7+1\/7*9+19*11+...+1\/1995*1997+1\/1997*1999的简便算 ...
观察数字，你会发现这样的规律：1\/1*3=（1-1\/3）*0.5 1\/3*5=（1\/3-1\/5）*0.5 1\/5*7=（1\/5-1\/9）*0.5 1\/7*9=（1\/7-1\/9）*0.5 ...那么1\/1*3+1\/3*5+1\/5*7+1\/7*9+19*11+...+1\/1995*1997+1\/1997*1999 =（（1-1\/3）+（1\/3-1\/5）+（1\/5-1\/9...
数学题1\/1*3+1\/3*5+1\/5*7+...1\/97*99+1\/99*101=?谢谢!
所以1\/3+1\/3*5+1\/5*7...1\/99*101 =[(1-1\/3)+(1\/3-1\/5)+...+(1\/99-1\/101)]\/2 =(1-1\/101)\/2 =50\/101 性质1 等式两边同时加上(或减去)同一个整式，等式仍然成立。若a=b 那么a+c=b+c 性质2 等式两边同时乘或除以同一个不为0的整式，等式仍然成立。若a=b 那么有a...
1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11的简便运算
解：令 1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11＝S 则 2S＝2\/1*3+2\/3*5+2\/5*7+2\/7*9+1\/9*11 ＝(1-1\/3)+(1\/3-1\/5)+(1\/5-1\/7)+(1\/7-1\/9)+(1\/9-1\/11)＝1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11 ＝1-1\/11 ＝10\/11 即原式＝S＝10...
简便运算:1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11
1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11 =1\/2*（1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11）=1\/2*（1-1\/11）=1\/2*10\/11 =5\/11
简便运算:1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11
1\/1*3+1\/3*5+1\/5*7+1\/7*9+1\/9*11 =1\/2*（1-1\/3+1\/3-1\/5+1\/5-1\/7+1\/7-1\/9+1\/9-1\/11）=1\/2*（1-1\/11）=1\/2*10\/11 =5\/11
