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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![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](/uploads/image/z/2679977-65-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2Can%3Da1%2B1%2F2a2%2B1%2F3a3%2B...%2B1%2Fn-1an-1%28n%3E1%29%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%2B1%3Da1%2B1%2F2a1%2B...%2B1%2Fnan+an%3Da1%2B1%2F2a2%2B...%2B1%2Fn-1an-1+%E4%B8%A4%E5%BC%8F%E7%9B%B8%E5%87%8F%E5%8F%AF%E5%BE%97+an%2B1-an%3D1%2Fnan+an%2B1%2Fan%3Dn%2B1%2Fn%E7%B4%AF%E4%B9%98%E5%8F%AF%E5%BE%97an%2Fa1%3Dn+%E6%89%80%E4%BB%A5an%3Dn%28n%3E%3D2%29n%3D1%E6%97%B6%E6%88%90%E7%AB%8B%2C%E6%89%80%E4%BB%A5an%3Dn%E6%AD%A3)
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
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正确答案是an=n/2,请问我那里做错了,想错了
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
累乘之后剩下的应该是
an/a2=(an/an-1)(an-1/an-2).(a3/a2)
=(n/n-1)(n-1/n-2).(3/2)
=n/2
你累乘的时候不能乘到a2/a1,因为n>1,明白了么?
an+1=a1+1/2a1+...+1/nan an=a1+1/2a2+...+1/n-1an-1 (n大于等于2) 两式相减可得 an+1-an=1/nan an+1/an=n+1/n(n大于等于2)累乘可得an+1/a2=n +1/2
2
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