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

\[2k-1\]

and inner radius \[2k-2\]

The k

^{th}annulus will have area

\[\pi ((2k-1)^2 - (2k-2)^2) = \pi ( 4k-3) \]

Adding up the areas of these

\[n\]

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

on using using these identities