So: an = 1/(3 n)

sn=a 1( 1-q^n)/( 1-q)=( 1/3)( 1- 1/3^n)/(2/3)=( 1- 1/3^n)/2 =( 1-an)/2

(2)bn = log(3)a 1+log(3)a2+…+log(3)an

=log(3)[a 1×a2×…×an]

=log(3)[ 1/3× 1/(3^2)×…× 1/(3^n)]

=log(3)[ 1/3^( 1+2+…+n)]

=-log(3)[3^( 1+2+…+n) ]

=-( 1+2+…+n)

=-n( 1+n) /2