In the arithmetic sequence an, the first term A1 = 1, the sequence BN = (1 / 2) an, and B1. B2. B3 = 1 / 64, we prove A1B1 + a2b2 +... + anbn

In the arithmetic sequence an, the first term A1 = 1, the sequence BN = (1 / 2) an, and B1. B2. B3 = 1 / 64, we prove A1B1 + a2b2 +... + anbn

If b1b2b3 is equal to 1 / 64, then B1 = 1 / 2A1 = 1 / 2, which is contradictory

a1b1+a2b2+…… Anbn = an for BN

Determinant = A1B1, A1B2, a1b3, I don't understand
Column 1 proposes B1
c2-b2c1,c3-b3c1
a1 0 0
a2 1 0
a3 0 2
How did this come about

Column 1 proposes B1
b1 a1 a1b2 a1b3
a2 a2b2+1 a2b3
a3 a3b2 a3b3+2
Multiply the first column by - B2 and add it to the second column. Multiply the first column by - B3 and add it to the third column
a1 0 0
a2 1 0
a3 0 2