若数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2)(n∈N*),求{an}的通项公式.
问题描述:
若数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2)(n∈N*),求{an}的通项公式.
答
∵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*),
两式相减,得nan=n(n+1)(n+2)-(n-1)n(n+1)(n∈N*),
∴an=3n+3.