(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 15:33:38
![(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?](/uploads/image/z/1782184-40-4.jpg?t=%EF%BC%881%2F2%2B1%2F3%E2%80%A6%2B1%2F2003%EF%BC%89%EF%BC%881%2B1%2F2%2B1%2F3%2B%E2%80%A6%2B1%2F2004%EF%BC%89-%EF%BC%881%2B1%2F2%2B1%2F3%2B%E2%80%A6%2B1%2F2003%EF%BC%89%EF%BC%881%2F2%2B1%2F3%2B%E2%80%A6%2B1%2F2004%EF%BC%89%3D%3F)
(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?
(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?
(1/2+1/3…+1/2003)(1+1/2+1/3+…+1/2004)-(1+1/2+1/3+…+1/2003)(1/2+1/3+…+1/2004)=?
设a=(1/2+1/3+...+1/2003) 原式=a(1+a+1/2004)-(1+a)(a+1/2004) =a+a^2+a/2004-a-1/2004-a^2-a/2004 =-1/2004