A1* a4 * a7 * ... a3n-2 = 2 (1+4+7+... 3n-2) = 2 [(3n-1) n/2] (sum by arithmetic progression).
log2 a 1+log2 a4+…log2 a3n-2 = log(2)a 1 * a4 * a3n-2=log(2)2^[(3n- 1)n/2]=.(3n- 1)n/2
Answer d