## Algebra 2 (1st Edition)

$a_n=\sum_{i=1}^n (2i-1) \pi$
Now, we have $a_1=\pi (1)^2 =\pi \\a_2=\pi (2)^2 -\pi =3\pi \\a_3=\pi (3)^2 -(2)^2 \pi =5\pi ....$ $a_n= a_1+(n-1)d$ and $a_n= \pi+(n-1) \times 2 \pi=(2n-1) \pi$ Thus, $a_i=(2i-1) \pi$ and $a_n=\sum_{i=1}^n (2i-1) \pi$