∵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.
若数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2)(n∈N*),求{an}的通项公式.
若数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2)(n∈N*),求{an}的通项公式.
数学人气:894 ℃时间:2019-09-24 04:45:45
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