# Solution to "I Would Walk Five Hundred Miles (Part 2)"

In the solution to Part 1 we found that Robert would walk $500e^{500}\approx 7.018\times 10^{219}$ miles.

Now let's see how far Derek would walk.

In the first quatrain Derek would walk

Then he would walk 799 miles for half of the additional miles. In other words he would walk $799\left( \frac{800^2}{2}\right)=\frac{799\cdot 800^2}{2}$ miles more, for a total of

Then he would walk 798 miles for a third of the last additional miles (798 times 1/3 of the last term). In other words, $798\cdot \frac{1}{3}\left(\frac{799\cdot 800^2}{2}\right)=\frac{798\cdot 799 \cdot 800^2}{2\cdot 3}$ miles more, for a total now of

Then adding 797 miles for a fourth of the last additional miles, we get a total of

We can see that eventually, the numerator will decrease to 0 times the previous additional miles ("Until I would walk 0 additional miles"). Factoring out a 800, we get a total distance of

(The next term would be when he reached 0 times the previous additional miles.)
The above can also be written as

Does that look familiar? That's right, it's the Choose function, also known as the binomial coefficient. Then this summation becomes

Notice that the sum of the numbers on any row of Pascal's triangle (values along row $k$ for $k\choose i$, starting at $k=0$) is a power of two:

This can be proven by the binomial expansion of $(1+1)^k$:

Therefore, how far Derek would walk can be written more succinctly.

So Derek would walk $800\cdot 2^{800}\approx 5.334\times 10^{243}$ miles.

Who gets the girl?

Therefore, Derek gets the girl and would walk $(800\cdot 2^{800}-500e^{500})$ miles more than Robert would walk.