已知数列{an}中各项都为正数,6Sn=an^2+3an+2,a1=2,求an.

Question

已知数列{an}中各项都为正数,6Sn=an^2+3an+2,a1=2,求an.

Answer 1

6Sn=an²+3an+2
6Sn-1=a(n-1)²+3a(n-1)+2
6Sn-6Sn-1=6an=an²-a(n-1)²+3an-3a(n-1)
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
数列各项都为正,an+a(n-1)>0,因此只有an-a(n-1)=3,为定值.
又已知a1=2,数列{an}是以2为首项,3为公差的等差数列.
an=2+3(n-1)=3n-1
n=1时,a1=3-1=2,同样满足.
数列{an}的通项公式为an=3n-1.