To travel a straight path of 10 meters in length, a flea uses the following strategy: each day it covers half the distance that was left from the previous day. Therefore, on the first day it covers 5 meters, on the second day 2.5 meters, and so on (the size of the flea is negligible).

(a) How many meters will it have traveled by the end of the seventh day? And by the tenth day?

(b) From which day will the flea be less than $0.001 \mathrm{~m}$ from the end of the path?
No $1^{\varrho}$ jump, the flea covers $10 \times \frac{1}{2}$, on the $2^{\underline{o}}$, $10 \times \frac{1}{2^{2}}$, and so on.

The jumps of the flea

![](https://cdn.mathpix.com/cropped/2024_05_01_8c8106cb3e7af661a7e3g-4.jpg?height=257&width=719&top_left_y=1028&top_left_x=680)

After 7 days, the flea will have covered

$$
\begin{aligned}
& 10\left(\frac{1}{2}\right)+10\left(\frac{1}{2}\right)^{2}+10\left(\frac{1}{2}\right)^{3}+10\left(\frac{1}{2}\right)^{4}+10\left(\frac{1}{2}\right)^{5}+10\left(\frac{1}{2}\right)^{6}+10\left(\frac{1}{2}\right)^{7}= \\
& 10\left[\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{4}+\left(\frac{1}{2}\right)^{5}+\left(\frac{1}{2}\right)^{6}+\left(\frac{1}{2}\right)^{7}\right]= \\
& 10 \times \frac{2^{6}+2^{5}+2^{4}+2^{3}+2^{2}+2+1}{2^{7}}=10 \times \frac{127}{128} \approx 9.9
\end{aligned}
$$

So in 7 days, she will have covered approximately $9.9 \mathrm{~m}$. In general, after $n$ days, the flea will have covered

$$
10\left(\frac{1}{2}+\cdots+\frac{1}{2^{n}}\right)
$$

To calculate the sum above, note that $\left(\frac{1}{2}+\cdots+\frac{1}{2^{n}}\right)$ is the sum of the $n$ terms of a geometric progression whose first term is $a_{1}=1 / 2$ and the common ratio is $q=1 / 2$. The formula for this sum is:

$$
S_{n}=\frac{a_{1}\left(1-q^{n}\right)}{1-q}=\frac{1 / 2\left(1-1 / 2^{n}\right)}{1-1 / 2}=1-\frac{1}{2^{n}}
$$

Therefore, $10\left(\frac{1}{2}+\cdots+\frac{1}{2^{n}}\right)=10\left(1-\frac{1}{2^{n}}\right)$.

Thus, at the end of the tenth day, the flea will have covered $10\left(1-\frac{1}{2^{10}}\right) \mathrm{~m}$.

The flea will be less than $0.001 \mathrm{~m}$ from the end of the path when she has already covered at least $10-0.001=9.999$, that is, when

$$
10\left(1-\frac{1}{2^{n}}\right) \geq 9.999
$$

Let's determine the smallest value of $n$ that satisfies the inequality above.

$$
\begin{aligned}
& 10\left(1-\frac{1}{2^{n}}\right) \geq 9.999 \quad \Longrightarrow 10-\frac{10}{2^{n}} \geq 9.999 \quad \Longrightarrow \quad 10-9.999 \geq \frac{10}{2^{n}} \\
& \Longrightarrow 0.001 \geq \frac{10}{2^{n}} \quad \Longrightarrow 2^{n} \geq \frac{10}{0.001} \quad \Longrightarrow 2^{n} \geq 10000
\end{aligned}
$$

Now,

$$
2^{10}=2^{5} \times 2^{5}=32 \times 32=1024
$$

it follows that

$$
2^{13}=2^{10} \times 2^{3}=1024 \times 8=8192
$$

Thus, we must have $n=14$.