(1)
因为(an,an+1)在函数f(x)=x^2+2x的图象上
an+1=an^2+2an
1+an+1=an^2+2an +1=(1+an)^2
lg(1+an+1)=2lg(1+an),（n>=1）,所以{lg(1+an)}等比
(2):已知a1=2,点（an,a(n+1)）在函数f(x)=x的平方+2x的图像上,
y=x^2+2x==>a(n+1)=an^2+2an
==>a(n+1)+1=an^2+2an+1
==>a(n+1)+1=(an+1)^2
∴当n=1,2,……n时有：
a2+1=(a1+1)^2
a3+1=(a2+1)^2=(a1+1)^4
a4+1=(a3+1)^2=(a1+1)^8
a5+1=(a4+1)^2=(a1+1)^16
………………
an+1=(a1+1)^(2^(n-1))=(2+1)^(2^(n-1))=3^(2^(n-1))
即：{an}的通项是：
an=3^(2^(n-1))-1
Tn=(1+a1)(1+a2)...(1+an)
==>(2+1)(3^2+1-1)(3^4+1-1).[3^(2^(n-1))+1-1]
==>3^(2(n-1))
(3)bn=(1/an)+[1/a（n+2）]
1/an=[1/(3^(2^(n-1))-1)]
1/a（n+2）=1/(3^(2^(n-1))+1
bn=(1/an)+[1/a（n+2）]=2*3^2^(n-1)/3^(2^n-1)
sn=b1+b2+b3+.+1/[3^(2^(n-1))-1]+1/[3^(2^(n-1)+1]
=1/2+1/4+1/8+.+1/[3^(2^(n-1))-1]+1/[3^(2^(n-1)+1]
=1/2+1/4+(1/2((1/2-1/4)+1/10+1/2(1/8-1/10)+.
∴2sn=1/2+1/4+1/8+1/10+.2/[(3^(2^(n-1))+1]
2sn=sn+1+1/[1/(3^(2^(n-1))-1)]+1/[(3^(2^(n-1))-1]
sn=1 -2/[3^(2^n)-1]
证明Sn+2/(3Tn-1)=1
Tn=3^(2^n -1)
3Tn=3^(2^n)
3Tn -1=3^(2^n) -1
2/(3Tn -1)=2/[3^(2^n) -1]
∴ Sn +2/(3Tn -1)=1 -2/[3^(2^n)-1] +2/[3^(2^n) -1]=1