已知数列{an}中,a1=2.a2=10 dm对任意n属于N*有a(n+2)=2a(n+1)+3an成立.(1)若{a(

a(n+2)=2a(n+1)+3an
a(n+2)- 3a(n+1) = - ( a(n+1) - 3an )
{a(n+1) - 3an} 是等比数列,q=-1
ie x=-3
a(n+1) - 3an = (-1)^(n-1) .( a2-a1)
= 8.(-1)^(n-1)
a(n+1) -2(-1)^(n+1) = 3[an - 2(-1)^n ]
{an - 2(-1)^n } 是等比数列,q=3
an - 2(-1)^n = 3^(n-1) ( a1 - 2(-1)^1 )
=4.3^(n-1)
an = 4.3^(n-1) + 2(-1)^n