已知数列{an}满足a1+2a2+3a3+...+nan=n(n+1)(n+2),则an= 过程详细点
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因为 a1+2a2+3a3+...+nan=n(n+1)(n+2),
所以 a1+2a2+3a3+...+(n-1)a(n-1)=(n-1)n(n+1),
两式相减,得 nan=n(n+1)[(n+2)-(n-1)]
所以 an=3(n+1)
======以下答案可供参考======
供参考答案1:
a1+2a2+3a3+...+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)
a1+2a2+3a3+...+nan=n(n+1)(n+2)
两式相减,得a(n+1)=3(n+1)(n+2)/(n+1)=3(n+2)
所以,an=3n+3
供参考答案2:
换角码把n换位n-1在做差