## Total Area of Concentric Annuli With Radii in Arithmetic Sequence

Suppose we have an arrangement of spherical concentric annuluses, each of thickness 1, separated by spaces of thickness 1.
The first annulus has outer thickness 1 and inner thickness 0. The second annulus has outer thickness 3 and inner thickness 2. The third annulus has outer thickness 5 and inner thickness 4. Continue in this way, then the k<supth annulus has outer radius

$2k-1$
$2k-2$
$\pi ((2k-1)^2 - (2k-2)^2) = \pi ( 4k-3)$
$n$
\begin{aligned} A &= \pi \sum^n_1 4k-3 \\ &= 4 \pi \sum^n_1 K - 3 \pi \sum 1 \\ &= 2 \pi n(n+1) - 3 \pi n \\ &= \pi n( 2n-1) \end{aligned}