已知数列{an}满足a1=1,an=a1 +1/2a2 +1/3a3 … +1/(n-1)a(n-1),(n>1,n∈N ),则an的通项公式a2=a1=1n>=3时an+1=a1+1/2a2+.+1/n-1an-1+1/nan两式相减得an+1-an=1/nan即an+1=n+1/nan即an+1/an=n+1/nan+1/a2=(an+1/an)(an/an-1).(a3/a2)=(n+1/n)(n/n-

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已知数列{an}满足a1=1,an=a1 +1/2a2 +1/3a3 … +1/(n-1)a(n-1),(n>1,n∈N ),则an的通项公式a2=a1=1n>=3时an+1=a1+1/2a2+.+1/n-1an-1+1/nan两式相减得an+1-an=1/nan即an+1=n+1/nan即an+1/an=n+1/nan+1/a2=(an+1/an)(an/an-1).(a3/a2)=(n+1/n)(n/n-

已知数列{an}满足a1=1,an=a1 +1/2a2 +1/3a3 … +1/(n-1)a(n-1),(n>1,n∈N ),则an的通项公式a2=a1=1n>=3时an+1=a1+1/2a2+.+1/n-1an-1+1/nan两式相减得an+1-an=1/nan即an+1=n+1/nan即an+1/an=n+1/nan+1/a2=(an+1/an)(an/an-1).(a3/a2)=(n+1/n)(n/n-
已知数列{an}满足a1=1,an=a1 +1/2a2 +1/3a3 … +1/(n-1)a(n-1),(n>1,n∈N ),则an的通项公式
a2=a1=1
n>=3时
an+1=a1+1/2a2+.+1/n-1an-1+1/nan
两式相减得an+1-an=1/nan
即an+1=n+1/nan
即an+1/an=n+1/n
an+1/a2=(an+1/an)(an/an-1).(a3/a2)
=(n+1/n)(n/n-1).(3/2)
=n+1/2
即an+1=n+1/2
即an=n/2
请问求得an+1/an=n+1/n后为什么不能直接用等比数列公式求an

已知数列{an}满足a1=1,an=a1 +1/2a2 +1/3a3 … +1/(n-1)a(n-1),(n>1,n∈N ),则an的通项公式a2=a1=1n>=3时an+1=a1+1/2a2+.+1/n-1an-1+1/nan两式相减得an+1-an=1/nan即an+1=n+1/nan即an+1/an=n+1/nan+1/a2=(an+1/an)(an/an-1).(a3/a2)=(n+1/n)(n/n-
等比数列是an+1/an=q是一个与一个与n无关的数

n+1/n又不是定值,怎么会是等比数列呢? 而且你这是从哪里搞来的解题过程?错的。n=1代入,a1=1/2,与已知不符。

因为比例系数不是常数