## Algebra 2 (1st Edition)

From part (b), we have $a_i=(2i-1) \pi$ and $a_n=\sum_{i=1}^n (2i-1) \pi$ For $n=1; a_1=\sum_{i=1}^1 (2i-1) \pi=\pi$ For $n=2; a_2=\sum_{i=1}^2 (2i-1) \pi=\pi+3 \pi=4 \pi$ For $n=4; a_4=\sum_{i=1}^4 (2i-1) \pi=\pi+3 \pi+5 \pi+7 \pi=16 \pi$ For $n=8; a_8=\sum_{i=1}^8 (2i-1) \pi=\pi+3 \pi+5 \pi+7 \pi+9 \pi+11 \pi+13 \pi+15 \pi=64 \pi$ When the number of rings is doubled, the total area is quadrupled.