(1/2-1/4)+(1/4-1/6)+…+(1/48-1/50)=( ) 为什么？

数学题(1\/2-1\/4)+(1\/4-1\/6)+(1\/6-1\/8)+……(1\/48-1\/50)=?
1\/2-1\/50=12\/25

(1\/2-1\/4)+(1\/4-1\/6)+(1\/6-1\/8)+……+(1\/48-1\/50)
=1\/2-1\/50 =25\/50-1\/50 =24\/50 =12\/25

1\/2x1\/4 1\/4x1\/6 …1\/48x1\/50
裂项法：1\/2x1\/4+1\/4x1\/6+ …+1\/48x1\/50 =1\/2×（1\/2-1\/4+1\/4-1\/6+……+1\/48-1\/50）=1\/2×（1\/2-1\/50）=1\/2×12\/25 =6\/25

2*4分之1加4*6分之1+6*8分之1···+48*50分之1=?
=1\/2×（1\/2-1\/4+1\/4-1\/6+……+1\/48-1\/50）=1\/2×（1\/2-1\/50）=1\/2×24\/50 =6\/25

计算,1\/2x4+1\/4x6+1\/6x8+...+1\/48x50=?,1\/2+1\/6+1\/12+1\/20+1\/30+1\/...
计算，1\/2x4+1\/4x6+1\/6x8+...+1\/48x50 =1\/2×（1\/2-1\/4+1\/4-1\/6+1\/6-1\/8+……+1\/48-1\/50）=1\/2×（1\/2-1\/50）=1\/2×12\/25 =6\/25，1\/2+1\/6+1\/12+1\/20+1\/30+1\/42 =1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6+1\/6-1\/7 =1-1\/7 =6...

简算1\/2*4+1\/4*6+1\/6*8+...1\/48*50
1\/(2*4)+1\/(4*6)+1\/(6*8)+……+(1\/(46*48)+1\/(48*50)=(1\/2)[(1\/2-1\/4)+(1\/4-1\/6)+(1\/6-1\/8)+……+(1\/46-1\/48)+(1\/48-1\/50)]=(1\/2)(1\/2-1\/50)=6\/25

【2×4】分之一加【4×6】分之一‘’‘一直加到【48×50】分之一
1\/(2x4)+1\/(4x6)+……+1\/(48x50) =(1\/2)x[(1\/2-1\/4)+(1\/4-1\/6)+……+(1\/48-1\/50)] =(1\/2)x(1\/2-1\/50) =1\/2x12\/25 =6\/25

数列计算 1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9=?
当n→∞时 1＋1\/2＋1\/3＋1\/4＋ … ＋1\/n 这个级数是发散的。简单的说，结果为∞ －－－ 用高中知识也是可以证明的，如下：1\/2≥1\/2 1\/3＋1\/4＞1\/2 1\/5＋1\/6＋1\/7＋1\/8＞1\/2 ……1\/[2^(k-1)+1]＋1\/[2^(k-1)+2]＋…＋1\/2^k＞[2^(k-1)](1\/2^k)＝1\/2...

1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9。。。+1\/100=?
1+1\/2+1\/3+...+1\/n≈lnn+C(C=0.57722...一个无理数,称作欧拉初始,专为调和级数所用)人们倾向于认为它没有一个简洁的求和公式.但是,不是因为它是发散的,才没有求和公式.相反的,例如等差数列是发散的,公比的绝对值大于1的等比数列也是发散的,它们都有求和公式.很高兴为您解答，祝你学习...

求(1-1\/2+1\/4-1\/6+1\/8-1\/10+……+1\/48-1\/50)*10的整数部分?
看不明白可以追问，求采纳，谢谢啦