Question

已知数列{An}的各项均为正数,前n项和Sn满足6Sn=An^2+3An+2

求通项公式...没有其它条件

Answer 1

因为A1=S1
==>6A1==A1^2+3A1+2，因为A1>0解得A1=1或2
因为6Sn=An^2+3An+2
...........①
当n≥2时有
6S<n-1>=A<n-1>^2+3A<n-1>+2...........②
由①-②可得6(Sn-S<n-1>)=An²-A<n-1>²+3An-3A<n-1>
因为Sn-S<n-1>=An
==>An²-A<n-1>²+3An-3A<n-1>=6An
即An²-A<n-1>²-3An-3A<n-1>=0
(An-A<n-1>)(An+A<n-1>)-3(An+A<n-1>)=0
(An+A<n-1>)(An-A<n-1>-3)=0
因为An>0
A<n-1>>0
所以An-A<n-1>-3=0==>An-A<n-1>=3
所以数列{An}是一个以1或2为首项,3为公差的等差数列
所以An=1+3(n-1)=3n-2
或An=2+3(n-1)=3n-1

Answer 2

6Sn=An^2+3An+2
6S(n-1)=[A(n-1)]^2+3A(n-1)+2
6Sn-6S(n-1)=6An=An^2+3An+2-{[A(n-1)]^2+3A(n-1)+2}
An-A(n-1)=3
{An}为等差数列，d=3
6S1=6A1=A1^2+3A1+2

A1^2-3A1+2=0
A1=1或2
An=1+3(n-1)=3n-2 或3n-1