Cops and robbers 2

Let $t_1$ be the time it takes to catch the first robber and take him to the police station. The overall metres travelled by the first robber doubled (because this doesn't include the trip back which is equally long) equals the metres travelled by the police at their first sprint. The first robber travels 60 seconds plus half of $t_1$ (as the other half goes driving back to the police station) with 5 m/s, and the police car travels $t_1$ seconds with 20 m/s.

$s=2\times5\times{(60+\frac{1}{2}t_1)}=20t_1$

$600+5t_1=20t_1$

$15t_1=600$

$t_1=40$

The second robber travels 60 seconds plus $t_1$ (which was spent catching the first robber) plus half $t_2$ (as the other half goes to the trip back to station) with 5 m/s. After doubled (because it doesn't include the trip back which is equally long), it equals the amount of metres travelled by the police car at the second sprint, which is 20 m/s times $t_2$ .

$2\times5\times{(60+t_1+\frac{1}{2}t_2)}=20t_2$

$600+10t_1+5t_2=20t_2$

$15t_2=600+10\times40$

$t_2=\frac{1000}{15}=\frac{5}{3}\times{t_1}$

From now on, for each ínteger $n$ where $n>1$ , we get the time spent to the $n$ th sprint (to catch the $n$ th robber and take him to the police station) = $t_n$ is

$20t_n=2\times5\times{(60+\sum_{k=1}^{n-1}(t_k)+\frac{1}{2}t_n)}$

$20t_n=600+10\sum_{k=1}^{n-1}(t_k)+5t_n$

$15t_n=600+10\sum_{k=1}^{n-1}(t_k)$

Which also means

$20t_{n+1}=2\times5\times{(60+\sum_{k=1}^{n}(t_k)+\frac{1}{2}t_{n+1})}$

$20t_{n+1}=600+10\sum_{k=1}^{n}(t_k)+5t_{n+1}$

$15t_{n+1}=600+10\sum_{k=1}^{n-1}(t_k)+10t_n=15t_n+10t_n$

$t_{n+1}=\frac{5}{3}\times{t_n}$

By looking at $t_1$ , $t_2$ and the result above, we see that the times $t_n$ form a geometric progression where $t_n=t_1\times(\frac{5}{3})^{n-1}=40\times(\frac{5}{3})^{n-1}$ . The total time spent is the sum of times spent for each robber:

$t=\sum_{i=1}^{20}(t_i)=\sum_{i=1}^{20}(40\times(\frac{5}{3})^{i-1})=\frac{40(1-(\frac{5}{3})^{20})}{1-\frac{5}{3}}=1641006.366753$

We get $\lfloor t\rfloor=\boxed{1641006}$ .