A1=2,递推公式A(n+1)=2An+n.(1)A1=2,A2=5,A3=12.(2)A(n+1)=2An+n,===>A(n+1)=2An+n.A(n+2)=2A(n+1)+(n+1).两式相减得：A(n+2)-A(n+1)+1=2[A(n+1)-An+1].===>A(n+1)-An+1=2^(n+1).(n=1,2,3,...).∴A2-A1+1=2^2,A3-A2+1=2^3,A4-A3+1=2^4,...An-A(n-1)+1=2^n.累加得：An-A1+(n-1)=2^(n+1)-2.===>An=[2^(n+1)]-n-1.经验证,其通项为An=[2^(n+1)]-n-1.(n=1,2,3,...).