1/(19×21)+1/(21×23)+1/(23×25)+……+1/(97×99)原式=1/2×(1/19-1/21)+1/2×(1/21-1/23)+1/2×(1/23-1/25)+……1/2×(1/97-1/99)=1/2×(1/19-1/21+1/21-1/23+1/23-1/25+1/25-1/27+……+1/95-1/97+1/97-1/99)=1/2(1/19-1/99)=40/1881我想知道一下
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![1/(19×21)+1/(21×23)+1/(23×25)+……+1/(97×99)原式=1/2×(1/19-1/21)+1/2×(1/21-1/23)+1/2×(1/23-1/25)+……1/2×(1/97-1/99)=1/2×(1/19-1/21+1/21-1/23+1/23-1/25+1/25-1/27+……+1/95-1/97+1/97-1/99)=1/2(1/19-1/99)=40/1881我想知道一下](/uploads/image/z/70072-16-2.jpg?t=1%2F%2819%C3%9721%29%2B1%2F%2821%C3%9723%29%2B1%2F%2823%C3%9725%EF%BC%89%2B%E2%80%A6%E2%80%A6%2B1%2F%2897%C3%9799%29%E5%8E%9F%E5%BC%8F%3D1%2F2%C3%97%281%2F19-1%2F21%29%2B1%2F2%C3%97%281%2F21-1%2F23%29%2B1%2F2%C3%97%281%2F23-1%2F25%29%2B%E2%80%A6%E2%80%A61%2F2%C3%97%281%2F97-1%2F99%29%3D1%2F2%C3%97%281%2F19-1%2F21%2B1%2F21-1%2F23%2B1%2F23-1%2F25%2B1%2F25-1%2F27%2B%E2%80%A6%E2%80%A6%2B1%2F95-1%2F97%2B1%2F97-1%2F99%EF%BC%89%3D1%2F2%281%2F19-1%2F99%29%3D40%2F1881%E6%88%91%E6%83%B3%E7%9F%A5%E9%81%93%E4%B8%80%E4%B8%8B)
1/(19×21)+1/(21×23)+1/(23×25)+……+1/(97×99)原式=1/2×(1/19-1/21)+1/2×(1/21-1/23)+1/2×(1/23-1/25)+……1/2×(1/97-1/99)=1/2×(1/19-1/21+1/21-1/23+1/23-1/25+1/25-1/27+……+1/95-1/97+1/97-1/99)=1/2(1/19-1/99)=40/1881我想知道一下
1/(19×21)+1/(21×23)+1/(23×25)+……+1/(97×99)
原式=1/2×(1/19-1/21)+1/2×(1/21-1/23)+1/2×(1/23-1/25)+……1/2×(1/97-1/99)
=1/2×(1/19-1/21+1/21-1/23+1/23-1/25+1/25-1/27+……+1/95-1/97+1/97-1/99)
=1/2(1/19-1/99)
=40/1881
我想知道一下原式的第一步和第二步怎么简化的呢?
1/(19×21)+1/(21×23)+1/(23×25)+……+1/(97×99)原式=1/2×(1/19-1/21)+1/2×(1/21-1/23)+1/2×(1/23-1/25)+……1/2×(1/97-1/99)=1/2×(1/19-1/21+1/21-1/23+1/23-1/25+1/25-1/27+……+1/95-1/97+1/97-1/99)=1/2(1/19-1/99)=40/1881我想知道一下
99-97=97-95=...21-19=2
1/(19×21)=1/(21-19)×(1/19-1/21)
任意1/(n×m)=1/(n-m)×(1/m-1/n)
易证
你不知道这种东西也没任何关系
1/(19×21)可以写成1/a×(1/19-1/21)是显然的
看题目可以发现必然要拆做两项差然后相消
这是这类题的思路
简单计算即可发现A为定值 故有所解
1/19-1/21=1/2*(21-19)/19*21=1/19*1/21,以后的式子都是这样,第二部把1/2提出来,然后括号里面相抵消就可以了
1/n(n+2)=(1/2)*2/n(n+2)=(1/2)*(1/n-1/(n+2))
1/((2n+1)*(2n+3))=1/2*(1/(2n+1)-1/(2n+3))