设数列{an}的前n项和为Sn,a1=1,Sn=Sn/n+2(n-1)

题目中Sn=Sn/n+2(n-1)应该是an=Sn/n+2(n-1)**
用a[n]表示第n项
1)a[n]=S[n]/n+2(n-1)
S[n]=na[n]-2n(n-1)
S[n-1]=(n-1)a[n-1]-2(n-1)(n-2)
当n≥2时两式相减:a[n]=S[n]-S[n-1]=na[n]-(n-1)a[n-1]-4(n-1)
整理可得:a[n]-a[n-1]=4
{a[n]}是以a[1]=1,d=4的等差数列
于是:a[n]=1+4(n-1)=4n-3
S[n]=n(a[1]+a[n])/2=2n^2-n.
2)1/(a[n]a[n+1])=(1/a[n]-1/a[n+1])/d=1/4*(1/a[n]-1/a[n+1])
T[n]=1/4*[1-1/5+1/5-…+1/(4n-3)-1/(4n+1)]=1/4[1-1/(4n+1)]＜1/4
又T[n]递增,T[n]≥T[1]=1/5
∴1/5≤Tn＜1/4.
3)S[n]/n=2n-1,S1/1=1,{S[n]/n}是等差数列,首项为1,公比为2
S[1]+S[2]/2+…+S[n]/n-(n-1)^2=n^2-(n-1)^2=2n-1=2009
∴n=1005.
**若S[n]=S[n]/n+2(n-1)
于是:(n-1)S[n]/n=2(n-1)
即:n≥2时,S[n]=2n
于是:a[1]=1,a[2]=3,a[n]=2,n≥3,这不是等差数列.