已知數列{an}滿足a1=1;an=a1+2a2+3a3+…+(n-1)a(n-1); (n>=2);求通項公式.(請給出過程,謝謝)

已知數列{an}滿足a1=1;an=a1+2a2+3a3+…+(n-1)a(n-1); (n>=2);求通項公式.(請給出過程,謝謝)


a2=a1+2a2=1+2a2得a2=-1an=a1+2a2+3a3+…+(n-2)a(n-2)+(n-1)a(n-1)a(n-1)=a1+2a2+3a3+…+(n-2)a(n-2)兩式相减:an-a(n-1)=(n-1)a(n-1)即an=na(n-1)所以an/a(n-1)=nan=[an/a(n-1)][a(n-1)/a(n-2)]…[a3/a2]a…



若數列{an}滿足a1+2a2+3a3+…+nan=n(n+1)(n+2)(n∈N*),求{an}的通項公式.


∵a1+2a2+3a3+…+nan=n(n+1)(n+2)(n∈N*),∴a1+2a2+3a3+…+(n-1)an-1=(n-1)n(n+1)(n∈N*),兩式相减,得nan=n(n+1)(n+2)-(n-1)n(n+1)(n∈N*),∴an=3n+3.