1.设bn等比为q,
b1=a1-2=-1
bn=-q^(n-1)
==>an=bn+2=2-q^(n-1)
2.a(n+1)=3/2an+1
令a(n+1)+t=3/2(an+t)
==>a(n+1)=3/2an+t/2
==>t=2
==>a(n+1)+2=3/2(an+2)
令bn=an+2,b1=a1+2=4
==>b(n+1)=3/2bn
==>bn是以4为首项,公比为3/2的等比数列
==>bn=4*(3/2)^(n-1)
==>an=bn-2=4*(3/2)^(n-1)-2