已知正项数列 an 的前n项和为sn,且满足a1=2,ana(n+ 1)=2(Sn +1) 求数

解：
a1a2=2(S1+1)=2(a1+1)
a1=2代入，
2a2=2(2+1)=6
a2=3
n≥2时，
ana(n+1)=2(Sn+1)=2Sn+2
a(n+1)a(n+2)=2S(n+1)+2
[2S(n+1)+2]-(2Sn+2)=2a(n+1)=a(n+1)a(n+2)-ana(n+1)
数列是正项数列，a(n+1)>0，等式两边同除以a(n+1)
a(n+2)-an=2，为定值
数列奇数项是以2为首项，2为公差的等差数列，偶数项是以3为首项，2为公差的等差数列
n为奇数时，an=2+(n-1)/2×2=n+1
n为偶数时，an=3+(n/2 -1)×2=n+1
综上，得数列的通项公式为an=n+1。

a1=1
n=1
a1.a2= 2(a1+1)
a2 =2(a1+1)/a1 = 4

an.a(n+1)=2(Sn +1)

Sn = (1/2)an.a(n+1) - 1

for n>=2
an = Sn - S(n-1)
=(1/2)an.a(n+1) - (1/2)a(n-1).an
1 =(1/2)a(n+1) - (1/2)a(n-1)
a(n+1) -a(n-1) =2
an - a(n-2) =2

if n is odd
an - a(n-2) =2
an - a1 = n-1
an = n

if n is even
an - a(n-2) =2
an - a2 =(n-2)
an =n+2

ie
an =n ; if n is odd
=n+2 ; if n is even