A racing turtle moves in a straight line as follows. In the first segment of the path, which measures $\frac{1}{2} \mathrm{~m}$, she runs at a speed of $3 \mathrm{~m} / \mathrm{s}$. In the second segment, which measures $\frac{1}{3} \mathrm{~m}$, she runs at a speed of $4 \mathrm{~m} / \mathrm{s}$. In the third segment, which measures $\frac{1}{4} \mathrm{~m}$, she runs at a speed of $5 \mathrm{~m} / \mathrm{s}$, and so on.

a) How long does it take the turtle to cover the first segment? Write the result as the difference of two unit fractions, that is, fractions with a numerator equal to 1.

b) Do the same with respect to the second segment.

c) Calculate the time it takes the turtle to cover the first 2013 segments.
Solution

a) Since speed is the ratio between the distance traveled and the time spent, we have, in the first segment,

$$
3=\frac{\frac{1}{2}}{t}
$$

and from this we obtain $t=\frac{1}{6} \mathrm{~s}$. To write $\frac{1}{6}$ as the difference of two unit fractions, it is enough to note that

$$
\frac{1}{6}=\frac{1}{2}-\frac{1}{3}
$$

b) Doing the same as in the previous item, we get

$$
4=\frac{\frac{1}{3}}{t}
$$

and from this we obtain $t=\frac{1}{4 \cdot 3} \mathrm{~s}=\frac{1}{12} \mathrm{~s}$. To write $\frac{1}{12}$ as the difference of two unit fractions, it is enough to note that

$$
\frac{1}{12}=\frac{1}{3 \cdot 4}=\frac{1}{3}-\frac{1}{4}
$$

c) The two previous items suggest that the times spent in each segment are, in order, $\frac{1}{2}-\frac{1}{3}, \frac{1}{3}-\frac{1}{4}, \frac{1}{5}-\frac{1}{4}, \ldots$ To see this, it is enough to note that

$$
\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}
$$

Thus, adding the times spent from the first to the $2013^{\text{th}}$ segment, we get

$$
\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots+\left(\frac{1}{2014}-\frac{1}{2015}\right)
$$

In this sum, many terms cancel out (the second of each parenthesis with the first of the next parenthesis). Thus, we obtain the sum of the time spent in the first 2013 terms as $\left(\frac{1}{2}-\frac{1}{2015}\right) \mathrm{s}$.