数列{an}中,A1=2 An+1=An+cn(c=2,n=1,2…）,且a1,a2,a3成公比不为1的等比数列 ,求{

a(n+1)=a(n)+2n=a(n)+n(n+1)-(n-1)n,
a(n+1)-n(n+1)=a(n)-(n-1)n,
{a(n)-(n-1)n}是首项为a(1)-0=2的常数数列.
a(n)-(n-1)n=2,
a(n)=2+(n-1)n,
[a(2)]^2=[2+1*2]^2=16,a(1)=2,a(3)=2+2*3=8,
[a(2)]^2=a(1)*a(3),满足条件.