(1-1/2-1/3-1/4…-1/2005)*(1/2+1/3+1/4+…1/2006)-(1-1/2-1/3-1/4-…-1/2006)*(1/2+1/3+1/4+…+1/2005)
来源:百度知道 编辑:UC知道 时间:2024/09/21 09:12:20
设A=1-1/2-1/3-1/4…-1/2005
B=1/2+1/3+1/4+…+1/2005
则原式=A*(B+1/2006)-(A-1/2006)*B
=A*B+A*1/2006-A*B+B*1/2006
=(A+B)*1/2006
=1*1/2006
=1/2006
换元法
可以设1-1/2-1/3...1/2005为a
1+1/2+1/3...1/2005为b然后带入
答案应该是1/1003
1/2006
(1+1/2+1/3+1/4)×
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
1+1/2+1+1/3+1+1/4+......+1/100=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)
1/1+1/2+1/3+1/4+。。。。+1/N 是多少
1/1+1/2+1/3+1/4+......1/2002=?
1-1/2+1/3-1/4+........1/99-1/100
求和Sn=1+(1+1/2)+(1+1/2+1/4)+....+[1+1/2+1/4.....+1/2^(n-1)]
数列 1+(1+1/2)+(1+1/2+1/4)+..............=?