A8. A tree grows as follows. On day 1, one branch grows from the ground. On day 2, a leaf grows on the branch, and the end of the branch splits into two new branches. On each subsequent day, a new leaf grows on each branch, and each end splits into two new branches. See the figure below.

How many leaves does the tree have at the end of the tenth day?
A) 172
B) 503
C) 920
D) 1013
E) 2047

![](https://cdn.mathpix.com/cropped/2024_04_17_998065f2d90ec501d25cg-2.jpg?height=166&width=714&top_left_y=2164&top_left_x=337)

day 1 day 2

![](https://cdn.mathpix.com/cropped/2024_04_17_998065f2d90ec501d25cg-2.jpg?height=440&width=574&top_left_y=1893&top_left_x=1164)

day 3

## B-questions
A8. D) 1013 The number of branch ends doubles every day. On days 1, 2, 3, ... there are 1, 2, 4, 8, ... new branches added. In general, on day $n$ exactly $2^{n-1}$ new branches are added.

The total number of branches on consecutive days is therefore $1, 1+2=3, 1+2+4=7, 1+2+4+8=15$, and so on. On day $n$, the total number of branches is $1+2+4+\cdots+2^{n-1}=2^{n}-1$.

The number of leaves grows each day by the number of branches from the previous day. On consecutive days, there are thus $0, 0+1, 0+1+3, 0+1+3+7, \ldots$ leaves. In general, the number of leaves on day $n$ is

$$
\begin{aligned}
\left(2^{0}-1\right)+\left(2^{1}-1\right)+\left(2^{2}-1\right)+\left(2^{3}-1\right)+\cdots+\left(2^{n-1}-1\right) & =\left(2^{0}+2^{1}+\cdots+2^{n-1}\right)-n \\
& =\left(2^{n}-1\right)-n .
\end{aligned}
$$

The total number of leaves at the end of day 10 is thus $2^{10}-11=1013$.