已知数列{an}满足a1=1,an=a1+1/2a2+1/3a3+...+1/n-1an-1(n>1)求数列{an}的通项公式an+1=a1+1/2a1+...+1/nan an=a1+1/2a2+...+1/n-1an-1 两式相减可得 an+1-an=1/nan an+1/an=n+1/n累乘可得an/a1=n 所以an=n(n>=2)n=1时成立,所以an=n正

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