tn=(a^(n- 1)bn)^n=a^(n^2-n)*bn^n

bn=tn/t(n- 1)=a^(2n-2)*bn^n/b(n- 1)^(n- 1)

[bn/b(n- 1)]^(n- 1)=( 1/a)^(2n-2]

For n >, bn/b (n-1) =1/a2; 2

2.

(∑ bi) 2 = ∑ bi 2+2 ∑ bi * bj (I<j)

Where ∑ bi * bj (I

Bi and bi^2 are both geometric series.

(∑bi)^2=[ 1/( 1- 1/a^2)]^2=a^4/(a^2- 1)^2

∑bi^2= 1/( 1- 1/a^4)=a^4/(a^4- 1)

∑bi * bj(I & lt； j)= 1/2[a^4/(a^2- 1)^2-a^4/(a^4- 1)]=a^4/2 * 2/(a^4- 1)

Therefore, {bn} = sum of products of all two different terms in ∑ bi * bj (i).