数列{an}满足a1=2,na(n+1)-3(n+1)an=-2n,则an=

na(n+1)-3(n+1)an=-2n²-4n-3
na(n+1)-n²-2n
=3(n+1)an-3n²-6n-3
=3(n+1)an-3(n+1)²
=3(n+1)(an-n-1)
即n[a(n+1)-(n+1)-1]=3(n+1)(an-n-1)
[a(n+1)-(n+1)-1]/(n+1)=3(an-n-1)/n
设数列｛bn｝,令bn=(an-n-1)/n
则有b(n+1)=3bn
所以｛bn｝是等比数列
又b1=a1-1-1=0
所以bn=b1*q^(n-1)=0
即(an-n-1)/n=0
所以an=n+1

a1=2,则a=2
代入得2n(n+1)-6n(n+1)=-2n,简化得(2n-6n)(n+1)=-2n,解得n=1/2,则2n=1