(1/2)+(1/3-1/2)+(1/4-1/3)...+(1/100-1/99)简便运算

来源：百度知道编辑：UC知道时间：2024/06/08 15:43:01

请写出过程

(1/2)+(1/3-1/2)+(1/4-1/3)...+(1/100-1/99)
=(1/2)+(-1/2+1/3)+(-1/3+1/4)...+(-1/99+1/100)
=[(1/2)-(1/2）]+[(1/3)-(1/3)]+[(1/4)-(1/4）]...[(1/99)-(1/99)]+1/100=
=1/100

其实就是后一个括号里面的减数和前一个括号的被减数相抵，这样到最后就等于1/100

好像没什么过程吧，每一项都约掉了，就剩下1/100了

(1/2)+(1/3-1/2)+(1/4-1/3)...+(1/100-1/99)
=(1/2)+(1/3)-(1/2)+(1/4)-(1/3)...+(1/99)-(1/98)+(1/100)-(1/99)
=1/100

把括号拆了后，就可以隔项相减。最后就只剩1/100了~~

(1+1/2+1/3+1/4)× (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+1/2+1+1/3+1+1/4+......+1/100=? (1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000) 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 (1/2) (1/4+1/6)= 数列 1+(1+1/2)+(1+1/2+1/4)+..............=?