(1+1/2+1/3+......+1/100)*(1/2+1/3+......+1/101)-(1+1/2+1/3+1/101)*(1/2+1/3+.
来源:百度知道 编辑:UC知道 时间:2024/05/29 00:55:25
求小学奥数解法
解:题目应该是求(1+1/2+1/3+……+1/100)*(1/2+1/3+……+1/101)-(1+1/2+1/3+……+1/101)*(1/2+1/3+……+1/100)的值
解:设1+1/2+1/3+……+1/100=A,1/2+1/3+……+1/101=B
所以B-A=1/101-1
因为1+1/2+1/3+……+1/101=1/2+1/3+……+1/101+1=B+1
1/2+1/3+……+1/100=1+1/2+1/3+……+1/100-1=A-1
所以原式=AB-(A-1)(B+1)
=AB-AB-A+B+1
=B-A+1
=1/101+1
=1/101
设1/2+1/3......1/100=A
(1+1/2+1/3......1/100)*(1/2+1/3......1/101)-(1+1/2+1/3......1/101)*(1/2+1/3+1/4+...+1/100)
=(1+A)*(A+1/101)-(1+A+1/101)*A
=A^2+102/101*A+1/101-A^2-A-1/101*A
=1/101
设1/2+1/3......1/101=A
则
(1+1/2+1/3......1/100)*(1/2+1/3......1/101)-(1+1/2+1/3......1/101)
=(1+A)*A-(1+A+1/101)
=A+A^2-1-A-1/101
=A^2-1-1/101
=(1/2+1/3......1/101)(1/2+1/3......1/101)-1-1/101
(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)+..............=?