n=1 a1=1*2*3=6n=2 a1+2a2=2*3*4=24 a2=9n=3 a1+2a2+3a3=3*4*5=60 a3=12 a1+2a2+3a3+...nan=n(n+1)(n+2)a1+2a2+3a3+...(n-1)*a(n-1)=(n-1)(n)(n+1) 相减nan=3n(n+1)an=3n+3存在一个等差数列
