(1)an=4n-3;(2)an=n2+n.
解:(1)∵an+1-an=[4(n+1)-3]-(4n-3)=4,
∴{an}为等差数列.
(2)由an=n2+n知a1=2,a2=6,a3=12,a2-a1≠a3-a2,
∴{an}不构成等差数列.