不明白
追答1/5*1/6=1/5-1/6
所以后面的一次类推··
又因为相邻两乘号中间所夹的所有数字可以以此类推的消除
即最后只剩下首项和末项 1/5-1/50
1\/5x1\/6+1\/6x1\/7+...1\/49x1\/50=?
原式等于1\/5-1\/6+1\/6-1\/7+···+1\/49-1\/50=1\/5-1\/50=9\/50
请教: 数学题1\/4x1\/5x1\/6+1\/5x1\/6x1\/7+1\/6x1\/7x1\/8+...+1\/98×1\/99...
1\/4*5*6+1\/5*6*7+1\/6*7*8+...+1\/98*99*100 =1\/2*{1\/4*5-1\/99*100} =247\/9900
数学题1\/4x1\/5x1\/6+1\/5x1\/6x1\/7+1\/6x1\/7x1\/8+...+1\/98×1\/99×1\/100...
同样得到1\/5x1\/6x1\/7=1\/2x(1\/5x1\/6-1\/6x1\/7)……1\/98×1\/99×1\/100=1\/2x(1\/98x1\/99-1\/99x1\/100)所以原式=1\/2x(1\/4x1\/5-1\/5x1\/6+1\/5x1\/6-1\/6x1\/7+……+1\/98x1\/99-1\/99x1\/100)=1\/2x(1\/4x1\/5-1\/99x1\/100)=1\/2x(1\/20-1\/9900)=1\/2x494\/9900=247\/9...
1\/3x1\/5+1\/5x1\/7+……+1\/97x1\/99
答案如图
1\/5x6十1\/7x8十1\/8x9十.…十1\/2024x2025等于多少?
=(1\/5-1\/6)+(1\/6-1\/7)+(1\/7-1\/8)+(1\/8-1\/9)+...+(1\/2023-1\/2024)+(1\/2024-1\/2025)=1\/5-1\/2025 =405\/2025 1\/5x6十1\/7x8十1\/8x9十.…十1\/2024x2025 =1\/5x6十1\/6x1\/7+1\/7x8十1\/8x9十.…十1\/2024x2025-1\/6x1\/7 =405\/2025-1\/6x1\/7 ...
一个问题???
应该是这样的，最后还剩：2000*（1-1\/2)(1-1\/3)(1-1\/4)...(1-1\/1999)(1-1\/2000)=2000*(1\/2)(2\/3)(3\/4)...(1998\/1999)(1999\/2000)=2000*(1\/2000)=1 所以最后还剩1个桃。
(1\/1x3)+(1\/3x5)+(1\/5x7)+……+(1\/49x51)=
(2n-1)(2n+1)] n≥1且n∈Z An= 1\/[(2n-1)(2n+1)]= 1\/2* [1\/(2n-1) -1\/(2n+1)]记数列An的前n项和为Sn。则：Sn=1\/2[1-1\/(2n+1)]，（累加相消）∴ Sn=(1\/1x3)+(1\/3x5)+(1\/5x7)+……+(1\/49x51)=1\/2[1-1\/(2*25+1)]=25\/51 ...
1.解方程。 2\/3X+1\/6X=1\/4 X-7\/9X=5\/12 3\/5X+20=50
干活
1\/6+1\/12+1\/20+1\/30+1\/42+1\/56=??? 紧急啊,当天说给分.
=1\/2×3+1\/3×4+1\/4×5+1\/5×6+1\/6×7+1\/7×8 =1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6+1\/6-1\/7+1\/8-1\/9 =1\/2-1\/9 =7\/18
1x1\/3+1\/3x1\/5+1\/5x1\/7+……1\/47x1\/49的简便运算
1x1\/3+1\/3x1\/5+1\/5x1\/7+……1\/47x1\/49 =2×（1-1\/3+1\/3-1\/5+1\/5-1\/7+……+1\/47-1\/49）=2×（1-1\/49）=96\/49 =1又49分之47
