1)sn=( 1/4)(an+ 1)^2

s(n- 1)=( 1/4)[a(n- 1)+ 1]^2

minus

And an=Sn-S(n- 1),

So 4an = (an+1) 2-[a (n-1)+1] 2.

[a(n- 1)+ 1]^2=(an+ 1)^2-4an=(an- 1)^2

A (n-1)+1= an-1or a (n-1)+1=-an+1.

If a (n-1)+1=-an+1.

a(n- 1)+a(n)=0

The contradiction between an and an is a positive integer sequence.

So a (n-1)+1= an-1.

an-a(n- 1)=2

So an is arithmetic progression and d=2.

a 1=S 1

So a1=1/4 (a1+1) 2.

(a 1- 1)^2=0

a 1= 1

an = 1+2(n- 1)= 2n- 1

2)

bn= 1/anxa(n+ 1)，

b 1= 1/ 1*3

b2= 1/3*5

.......

bn = 1/((2n- 1)*(2n+ 1))

TN = 1/ 1 * 3+ 1/3 * 5+....+ 1/(2n- 1)(2n+ 1)

= 1/2*( 1- 1/3)+ 1/2*( 1/3- 1/5)+....+ 1/2 *[ 1/(2n- 1)- 1/(2n+ 1)]

= 1/2[( 1- 1/3)+( 1/3- 1/5)+....+ 1/(2n- 1)- 1/(2n+ 1)]

= 1/2 *[ 1- 1/(2n+ 1)]

∫ 1/(2n+ 1)>0,∴ 1- 1/(2n+ 1)<； 1,∴ 1/2*[ 1- 1/(2n+ 1)]<； 1/2. That's TN.