第1个回答 2011-01-24
a[n]=S[n]/n+2(n-1)
na[n]=S[n]+2n(n-1)
(n-1)a[n]=S[n]-a[n]+2n(n-1)=S[n-1]+2n(n-1)
a[n]=S[n-1]/(n-1)+2n ------------(1)
同时因为a[n]=S[n]/n+2(n-1)
有a[n-1]=S[n-1]/(n-1)+2(n-2) ------(2)
(1)-(2),得
a[n]-a[n-1]=4
所以{an}是等差数列,且公差为4,这样
a[n]=a[1]+4*(n-1)=4n-3