a1+2a2+3a3……nan=n（n+1）*（n+2）求通项公式

a1+2a2+...+nan=n(n+1)(n+2)
a1+2a2+..+(n-1)a(n-1)=(n-1)n(n+1)
两式相减:
nan=n((n+1)(n+2)-(n-1)(n+1))
an=n^2+3n+2-n^2+1
=3n+3
a1=1*2*3=6也满足此式
所以an通项公式an=3n+3,n∈N

额
a1+2a2+3a3...+nan=n*(n+1)*(n+2) (1)
a1+2a2+3a3...+nan+(n+1)a(n+1)=(n+1)*{(n+1)+1}*[(n+1)+2] (2)
式(2)-(1)整理一下即可得出
a(n+1)=3(n+2)=3[(n+1)+1]
所以an=3(n+1)

一楼的对了

2n(n+1)

an=3(n+1)