a1+2a2+3a3+~+nan=n(n+1)(2n+1)知,
a1+2a2+3a3+~+(n-1)an-1=(n-1)n(2n-1),
n≠1时两式相减知an=(n+1)(2n+1)-(n-1)(2n-1)=6n,
n=1时a1=6,满足an=6n
所以an=6n