数列题,十万火急

题目

数列题,十万火急
已知正项数列（An)满足2Sn=An^2+An-30,n属于N+,数列（bn)满足Bn=2^(An-5),求（1）分别求（An),(Bn)通项公式（2）求证（Bn）是等比数列

答案

把n=n+1代入得2Sn+1=a(n+1)^2+a(n+1)-30,用上式减原式得：2a(n+1)=a(n+1)^2-an^2+a(n+1)-an,整理得：an^2+an+1/4=a(n+1)^2-a(n+1)+1/4,继续整理得：(an+1/2)^2=[a(n+1)-1/2]^2,两边开根得：an+1=a(n+1),把n=1代入原...