在正整数数列中前n项和Sn满足Sn=1/8*(an+2)^2求证是等差数列,若Cn=1/（an*an+1）,求Cn的前n项和Tn_数学解答

在正整数数列中前n项和Sn满足Sn=1/8*(an+2)^2求证是等差数列,若Cn=1/（an*an+1）,求Cn的前n项和Tn.

人气：447 ℃　时间：2020-05-09 20:00:12

解答

Sn=1/8*(An+2)^2有Sn-1=1/8*(An-1+2)^2相减;an=1/8*(An+2)^2-1/8*(An-1+2)^2整理得（An-An-1-4）*(An+An+1)=0正整数数列An-An-1=4An等差数列Sn=1/8*(An+2)^2有a1=1/8*(a1+2)^2a1=2An=4n-2Sn=2n^2Cn=1/（an*an+1）=1/...