Trevor rides Jake piggyback I

First, define a "step" as one north-east-south-west travel done successively. It is obvious that travelling north/south increases/decreases $y$ -coordinate and travelling east/west increases/decreases $x$ -coordinate.

Let the symbol $\to$ denote the transition from one step to the other and $(a,b)\to (c,d)$ means that Trev-Jake displaced from $(a,b)$ to $(c,d)$ after a step. From the given data, we have,

$(0,0)\to \left(\frac12-\frac14,1-\frac13\right)\to \left(\frac12-\frac14+\frac16-\frac18,1-\frac13+\frac15-\frac17\right)\to \ldots$

As the steps continue ad infinitum, the limiting displacement point can be given as,

$(X,Y)=\left(\sum_{n=0}^\infty\frac{(-1)^{n}}{2(n+1)}~,~\sum_{n=0}^\infty\frac{(-1)^{n}}{2n+1}\right)$

We need to compute $X$ and $Y$ . Now, by the Maclaurin series of $\arctan(x)$ and $\ln(1+x)$ , we can obtain that,

$X=\sum_{n=0}^\infty\frac{(-1)^{n}}{2(n+1)}=\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^{n}}{n+1}=\frac12\ln(1+1)=\frac12\ln2\\ Y=\sum_{n=0}^\infty\frac{(-1)^{n}}{2n+1}=\arctan(1)=\frac{\pi}{4}$

Using the distance formula, we can easily find the displacement as follows:

$\begin{aligned}\textrm{Displacement}(s)&=\sqrt{(X-0)^2+(Y-0)^2}\\&=\sqrt{X^2+Y^2}\\&=\sqrt{\frac{1}{4}\ln^22+\frac{\pi^2}{16}}=\frac12\sqrt{\ln^22+\frac{\pi^2}{4}}\end{aligned}$

Comparing values with the given form in question gives us the answer as $2+4=\boxed{6}$ .