已知数列an,an>0,Sn=a1+a2+a3.+an,且an=6Sn/an + 3,求Sn!已知数列an,an>0,Sn=a1+_数学解答

已知数列an,an>0,Sn=a1+a2+a3.+an,且an=6Sn/an + 3,求Sn!
已知数列an,an>0,Sn=a1+a2+a3.+an,且an=6Sn/an + 3〔不是a（n+3）,是an + 3,〕求Sn!

人气：389 ℃　时间：2020-02-03 02:09:45

解答

An=6Sn/(An+3)6Sn=(An)^2+3Ann>=26S(n-1)=(A(n-1))^2+3A(n-1)6An=(An)^2+3An-(A(n-1))^2-3A(n-1)(An)^2-(A(n-1))^2=3An+3A(n-1)(An+A(n-1))(An-A(n-1))=3(An+A(n-1))An>0An+A(n-1)≠0An-A(n-1)=36A1=6S1=(A1)^2+3A1A...