1/()+1/()+1/()+1/()+1/()+1/()+1/()=1
来源:百度知道 编辑:UC知道 时间:2024/06/25 08:20:28
1/()+1/()+1/()+1/()+1/()+1/()=1
1=1/2+1/4+1/8+1/16+1/24+1/48
1=1/2+1/4+1/8+1/16+1/24+1/48
这是什么问题 括号里面全填6不行么
1=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6
所以1-1/2=1/2
1/2-1/3=1/6
1/3-1/4=1/12
1/4-1/5=1/20
1/5-1/6=1/30
1/6
1=1/2+1/6+1/12+1/20+1/30+1/6
其实这是极限求和的一种吧
1=1/2+1/4+1/8+1/16+1/32+1/64+1/64
1/(7)+1/(7)+1/(7)+1/(7)+1/(7)+1/(7)+1/(7)=1
1/7+1/7+1/7+1/7+1/7+1/7+1/7=1
1/2+1/4+1/8+1/16+1/24+1/48=1
1/96+1/48+1/32+1/16+1/8+1/4+1/2=1
1/54+1/36+1/27+1/18+1/12+1/9+1/3+1/3=1
..................
太多了!时间有限,就写这么多把!
1/8=1/( )+1/( )1 1/10=1/( )+1/( )1/12=1/( )+1/( )+1/( )
1/8=1/()+1/()
(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)*……*(1-1/99)*(1+1/99)。
计算1+1/2+(1/3-1)+1/4+1/5+(1/6-1/2)+1/7+1/8+(1/9/1/3)......
(2/1+3/1+4/1+5/1)乘(2/1+3/1+4/1+5/1+6/1)减(2/1+3/1+4/1+5/1+6/1)乘(2/1+3/1+4/1+5/1)等于多少
1-(1-1/2)-(1/2-1/3)-…-(1/2004-1/2005)
1+1/(2-1/3)/【1-1/(2+1/3)】
(1/2005-1)(1/2004-1)...(1/3-1)(1/2-1)
数列求和:1+(1+1/2)+、、、+(1+1/2+1/3+、、、+1/n)
(9-1/2)×(1+1/2)×(1-1/3)×(1-1/4)×(1+1/4)×┈×(1-1/2005)×(1-1/2006)怎么做??