1/(1*3)+1/(3*5)+1/(5*7)+……+1/(19*21)=?
来源:百度知道 编辑:UC知道 时间:2024/05/14 19:50:38
请说明简便算法
裂项法:1/(2n-1)(2n+1)=1/2[1/(2n-1)-1/(2n+1)]
原式=1/2*(1/1-1/3)+1/2*(1/3-1/5)……1/2*(1/19-1/21)
=1/2*[(1-1/3+1/3-1/5……+1/19-1/21)]
=1/2*(1-1/21)
=1/2*20/21
=10/21
原式=1/2×[(1/1-1/3)+(1/3-1/5)+……+(1/19-1/21)]
=1/2×(1—1/21)=10/21
1/(1*3)+1/(3*5)+1/(5*7)+……+1/(19*21)=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+……1/2(1/19-1/21)=1/2(1-1/3+1/3-1/5+1/5-1/7+……1/19-1/21)=1/2(1-1/21)=10/21
把每一项都拆开就是了!例如:1/(3*5)可以拆成两项1/3-1/5,把每一项 都这样拆开,答案就很明显了!就不多说了
因为1/(1*3)=1/(3-1)*(1-1/3)=1/2*(1-1/3)
1/(3*5)=1/(5-3)*(1/3-1/5)=1/2*(1/3-1/5)
……………………………
所以1/(19*21)=1/(21-19)*(1/19-1/21)=1/2*(1/19-1/21)
所以原式=1/2(1-1/3+1/3-1/5+1/5-……-1/19+1/19-1/21)
=1/2(1-1/21)
=1/2*20/21
=10/21
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
(1-1/100)(1-1/99)(1-1/98)......(1-1/3
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)
1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100)
1+1/2+1+1/3+1+1/4+......+1/100=?
1+1/2+1/3+.....+1/n
1+1/2+1/3+...+1/100
1+1/3+1/6+........+1/55
1-1/2+1/3-.....-1/10