Rocky's Run

This is a generalized solution for the problem.

Let $f(x)$ denote the number of people that started running from the $x^{\textrm{th}}$ mile. Suppose that the running population gets multiplied by $a$ after every mile. We have the following recurrence:

$f(x+1)=af(x)~,~f(0)=1 \implies f(x)=a^x$

Since the multiplying of people happens after every single mile, we can say that the total miles ran by the people from the $k^{\textrm{th}}$ mile to the $(k+1)^{\textrm{th}}$ mile is equal to $f(k)$ .

Let the total amount of miles ran by all when they reach the $k^{\textrm{th}}$ mile be $D(k)$ . We then have,

$D(k)=\sum_{i=0}^{k-1} f(i)\\ \implies D(k) = \sum_{i=0}^{k-1} (a^i)$

The formula for summation of a finite GP comes to the rescue. We get,

$\boxed{D(k)=\dfrac{a^k-1}{a-1}}$

For the following problem, we have $a=2$ (since doubling takes place) and $k=10$ (since Rocky runs 10 miles).

Hence, $D(10)=\dfrac{2^{10}-1}{2-1}=\boxed{1023}$

He runs 10 miles, so the the total sum is $2^0+2^1+…+2^9$ . Note that this does not include $2^{10}$ because 0-10 is a total of 11 miles instead of 10. This sum is equal to 2^10-1 = 1023

After Rocky will run 1 mile 1 person will join him. After running another mile the no. of persons will be the double i.e. 2 person will join him. Then after running one more mile the no. of people joining him will become 4. This pattern will continue and

Thus, combined number of miles ran by everyone = $10+9 \times 1+8 \times 2+7 \times 4+6 \times 8+5 \times 16+4 \times 32+3 \times 64+2 \times 128+1 \times 256$

Therefore, the answer is $\huge 1023$ $miles$

Let $N$ be the number of miles Rocky runs.

Let $M$ be the total miles ran by the whole crowd.

Let $C$ be the size of the crowd at the finish after running $N$ miles. ( The crowd doubles as they arrive at the finish line. )

$C = 2^{N}$ (If you want to calculate the crowd just before the finish, use $2^{N-1}$ )

$M = C - 1$

If Rocky runs 10 miles, that makes $M = 2^{10} - 1$ , and so the answer is $\boxed{1023}$ miles.

For fun, here's a Python simulation, too:

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crowd = 1
total_miles = 0
rocky_miles = 0
while rocky_miles < 10:
    total_miles += crowd
    rocky_miles += 1
    crowd *= 2
print "Answer:", total_miles

$2^{0} \rightarrow 2^{10}$ (useful for visualization):

powers of 2