酒店浴巾花样叠法:归纳 猜想 证明

来源:百度文库 编辑:杭州交通信息网 时间:2024/04/28 18:03:36
数列的前4项为:
1/1,1/1+1/(1+2),1/1+1/(1+2)+1/(1+2+3),1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)…,猜想通项公式an.

因为1+2+3+...+n=n(n+1)/2

故:1/(1+2+3+...+n)=2/n(n+1)=2[1/n-1/(n+1)]

an=1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+...+n)

=2[1/1-1/2]+2[1/2-1/3]+...+2[1/n-1/(n+1)]

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

=2[1-1/(n+1)]

=2n/(n+1)

因为1+2+3+...+n=n(n+1)/2

故:1/(1+2+3+...+n)=2/n(n+1)=2[1/n-1/(n+1)]

an=1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+...+n)

=2[1/1-1/2]+2[1/2-1/3]+...+2[1/n-1/(n+1)]

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

=2[1-1/(n+1)]

=2n/(n+1)