Question

a1+2a2+3a3……nan=n（n+1）*（n+2）求通项公式

Answer 1

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