## Task A-1.4.

In the plane, one hundred circles with the same center and radii $1,2, \ldots, 100$ are drawn. The smallest circle is colored red, and each of the 99 circular rings bounded by two circles is colored either red or green such that adjacent areas are of different colors.

Determine the total area of the green-colored regions.
## Second Solution.

As in the first solution, we conclude that the area of the green regions is

$$
P_{z}=\left(2^{2} \pi-1^{2} \pi\right)+\left(4^{2} \pi-3^{2} \pi\right)+\cdots+\left(100^{2} \pi-99^{2} \pi\right)
$$

Since $n^{2} \pi-(n-1)^{2} \pi=(2 n-1) \pi$, we get

$$
P_{z}=3 \pi+7 \pi+11 \pi+\cdots+195 \pi+199 \pi
$$

Let's write this sum twice (once in ascending order, and once in reverse) and add the resulting equations:

$$
\begin{array}{cccccc}
P_{z}=3 \pi+7 \pi & +\cdots & +195 \pi & +199 \pi \\
P_{z}=199 \pi+195 \pi+\cdots & +\cdots \pi & + & + & \\
\hline 2 P_{z} & =202 \pi+202 \pi+\cdots & +202 \pi & +202 \pi
\end{array}
$$

It follows that $2 P_{z}=50 \cdot 202 \pi$, i.e., $P_{z}=5050 \pi$.

Note: The correct answer without justification earns 1 point.

Note: If the student calculates the sum of the areas of the red regions (e.g., because they are confused about the order of the red and green regions), they should receive 1 point less than they would have if they had calculated the sum of the areas of the green regions using the same method.