1/n(n+ 1)=n-Sn-an

bn=an+ 1/n(n+ 1)

bn=an+n-Sn-an

bn=n-Sn

b(n- 1)=(n- 1)-(Sn- 1)

b(n- 1)=n-Sn

b(n- 1) : bn= 1

And a1=1/4b1= 3/4.

Bn is b1= 3/4; Geometric series of q= 1

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

an = b 1-(n+ 1)-n/n(n+ 1)

an = b 1-( 1/n)+ 1/(n+ 1)

a(n- 1)= b 1-[ 1/(n- 1)]+ 1/(n)

a 1 =b 1- 1+(.......)

sn = nb 1- 1+ 1/(n+ 1)

because

nan = nb 1- 1/(n+ 1)

therefore

Cn=2nb 1- 1

Cn = (3n/2)- 1

c(n- 1)= 3(n- 1)/2- 1

Obviously, Cn is arithmetic progression.

C 1= 1/2

TN = n[(3n/2)- 1/2]/2 =(3n- 1)n/4 & gt； 0

TN < 1 is invalid. . .

What is Tn? . .