2011 AIME II 第 5 题

Terms $2012$ through $4022$ are $r^{2011}$ times the first $2011$ terms

Each successive block of $2011$ terms is the previous block times $r^{2011};$ find that ratio from the two given sums

Group the series into blocks of $2011$ consecutive terms. Each term of the second block is $r^{2011}$ times the corresponding term of the first block, so the block sums form a geometric sequence with ratio $r^{2011}.$ The first block sums to $200$ and the second block sums to $380 - 200 = 180,$ so $r^{2011} = \frac{180}{200} = \frac{9}{10}.$

The third block then sums to $180 \cdot \frac{9}{10} = 162,$ so the sum of the first $6033$ terms is $380 + 162 = 542.$