已知数列An+1=An+3n+1,A1=1,求An的通项公式

a(n+1)=a(n)+3n+1 = a(n) + (3/2)[n(n+1)-(n-1)n] + [(n+1)-n],a(n+1) - 3n(n+1)/2 - (n+1) = a(n) - 3(n-1)n/2 - n,{a(n)-3(n-1)n/2 - n}是首项为a(1)-0-1=0的常数数列.a(n)-3(n-1)n/2 - n = 0,a(n) = 3(n-1)n/2 + ...