a1+2a2+3a3+...+(n-1)a(n-1)+nan=n(n+1)(n+2)a1+2a2+3a3+...+(n-1)a(n-1) =(n-1)n(n+1)两式相减an=3(n+1),an是等差数列Sn=(a1+an)*n/2=[6+3(n+1)]*n/2=(3n+9)*n/2
