若{an}为正项数列.Sn为其前N项和,且an,Sn,an^2成等差数列’ 1.求{an},2,设f(n)＝Sn/（n+

an,Sn,an^2成等差数列
2Sn=an^2+an
2Sn-1=an-1^2+an-1 相减
2an=an^2+an-(an-1^2+an-1)
(an+an-1)=(an-an-1)(an+an-1) 若{an}为正项数列
an-an-1=1
an=an-1+1
{an}为 首项是 a1=1 d=1的等差数列
an=a1+(n-1)d=n
Sn=n(n+1)/2
2.f(n)=(n(n+1)/2)/(n+50)*(n+1)(n+2)/2
=n/(n+50)(n+2)
=n/[n^2+52n+100]
=1/[n+100/n+52]
n+100/n+52>=20+52 均值定理
n+100/n+52最小值=72
f（n）的最大值=1/72

a1+a1^2=2*a1,a1>0,a1=1,a2+a2^2=2*(1+a2),a2>0,a2=2,同样a3=3，推测an=n,严格的话用数学归纳法证即可知an=n，简单点直接代入那个关系。
2.按分母是（n+50)Sn+1理解，f（n）=1/(n+(100/n)+52),转化求n+(100/n)的极小值，用a^2+b^2>=2ab,a,b>=0,f(n)最大值1/72