(1/2+1/3+1/4+……+1/100)x(1+1/2+1/3+……+1/99)-(1/2+1/3+……+1/99)(1+1/2+1/3+……+1/100)
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![(1/2+1/3+1/4+……+1/100)x(1+1/2+1/3+……+1/99)-(1/2+1/3+……+1/99)(1+1/2+1/3+……+1/100)](/uploads/image/z/3752999-71-9.jpg?t=%EF%BC%881%2F2%2B1%2F3%2B1%2F4%2B%E2%80%A6%E2%80%A6%2B1%2F100%EF%BC%89x%EF%BC%881%2B1%2F2%2B1%2F3%2B%E2%80%A6%E2%80%A6%2B1%2F99%EF%BC%89-%EF%BC%881%2F2%2B1%2F3%2B%E2%80%A6%E2%80%A6%2B1%2F99%EF%BC%89%EF%BC%881%2B1%2F2%2B1%2F3%2B%E2%80%A6%E2%80%A6%2B1%2F100%EF%BC%89)
(1/2+1/3+1/4+……+1/100)x(1+1/2+1/3+……+1/99)-(1/2+1/3+……+1/99)(1+1/2+1/3+……+1/100)
(1/2+1/3+1/4+……+1/100)x(1+1/2+1/3+……+1/99)-(1/2+1/3+……+1/99)(1+1/2+1/3+……+1/100)
(1/2+1/3+1/4+……+1/100)x(1+1/2+1/3+……+1/99)-(1/2+1/3+……+1/99)(1+1/2+1/3+……+1/100)
(1/2+1/3+1/4+……+1/100)x(1+1/2+1/3+……+1/99)-(1/2+1/3+……+1/99)(1+1/2+1/3+……+1/100)
令a=1/2+1/3+……+1/99
原式=(a+1/100)(1+a)-(a)(1+a+1/100)
=a+a^2+1/100+a/100-a-a^2-a/100
=1/100
有不懂欢迎追问
解答如下:
原式=(1/2+1/3+......+1/100)x(1+1/2+......+1/99)-(1/2+1/3+......+1/99)x(1+1/2+......+1/99)
-(1/100)x(1/2+1/3+......+1/99) 提出公因式(1/2+1/3+......+1/99)
=(1+1/2+1/3+......+1/99)x(1/100)-(1/100)x(1/2+1/3+......+1/99)
=1/100
设几个字母表示看起来清楚些:
设A=1/2+1/3......+1/100,B=1+1/2+1/3+.....1/99,C=1/2+1/3+......1/99,D=1+1/2+1/3+......+1/100。
原式=AB-CD
=(C+1/100)B-C(B+1/100)
=CB+1/100B-CB-1/100C
=1/100B-1/100C
=1/100(B-C)
=1/100