The diagram shows the first four levels of a school's emergency telephone tree. In the case of an emergency, the principal calls two students. The first level consists of just the principal, and in the second level, two students are contacted. In the next level, each of these two students contacts two students who have not been contacted, so after the third level, a total of 6 students have been contacted. This continues so that each student contacts two students who have not yet been contacted. After the $8^{\text {th }}$ level, how many students in total have been contacted?

![](https://cdn.mathpix.com/cropped/2024_04_17_d0cbae19e125c7a52478g-2.jpg?height=659&width=1203&top_left_y=478&top_left_x=518)
The number of students contacted in level 1 is 0 .

The number of students contacted in level 2 is $2=2^{1}$.

The number of students contacted in level 3 is $4=2^{2}$.

The number of students contacted in level $n$ is twice the number of students that were contacted in the previous level.

This means the number of students contacted in level 4 is $2\left(2^{2}\right)=2^{3}$, the number of students contacted in level 5 is $2\left(2^{3}\right)=2^{4}$, and so on.

The number of students contacted after 8 levels is therefore

$$
\begin{aligned}
2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}+2^{7} & =2+4+8+16+32+64+128 \\
& =254 .
\end{aligned}
$$

#