已知数列{An}的前n项和为Sn,点(n,Sn) (n属于N+）在函数f(x)=3x^2-2x图象上

题目

已知数列{An}的前n项和为Sn,点(n,Sn) (n属于N+）在函数f(x)=3x^2-2x图象上
(1)求数列{An}的通项公式An
(2)设bn=3/An*A(n+1) 数列{bn}的前n项和记为Tn,求使得|Tn-1/2|＜1/100 成立的最小整数n

答案

Sn=3n^2-2nSn-1=3(n-1)^2-2(n-1)an=6n-5bn=3/(6n-5)(6n+1)=1/2(1/(6n-5)-1/(6n+1))tn=1/2(1/a1-1/(6n+1))=1/2(1-1/(6n+1))=3n/(6n+1)tn-1/2=1/(6n+1)100 n>33/2 n=34