## 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 red or green such that adjacent areas are of different colors.

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

If the radius of the larger circle is $n$, and the smaller circle is $n-1$, the area of the circular ring between them is $n^{2} \pi-(n-1)^{2} \pi$.

Since every other area is green, the total area of the green regions is

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

If we apply the formula for the difference of squares $n^{2}-(n-1)^{2}=1 \cdot(n-1+n)$, we get

$$
P_{z}=(1+2+3+\cdots+99+100) \pi
$$

According to the formula for the sum of the first $n$ natural numbers, $1+2+\cdots+n=\frac{n(n+1)}{2}$, we get $P_{z}=\frac{100 \cdot 101 \pi}{2}=5050 \pi$.

Note: Students who do not know the formula for the sum of the first $n$ natural numbers can complete the solution using Gauss's insight. Write the sum twice and add the resulting equations as follows:

$$
\begin{array}{cccccccccc}
P_{z}= & \pi & + & 2 \pi & + & \cdots & + & 99 \pi & + & 100 \pi \\
P_{z}= & 100 \pi & + & 99 \pi & + & \cdots & + & 2 \pi & + & \pi \\
\hline 2 P_{z}= & 101 \pi & +101 \pi & + & \cdots & + & 101 \pi & + & 101 \pi
\end{array}
$$

It follows that $2 P_{z}=100 \cdot 101 \pi$, i.e., $P_{z}=5050 \pi$.