Trevor rides Jake piggyback II

Since we require the distance of the final position from the origin $(0,0)$ , we can assume without loss of generality that the first step is upwards, that is, in the positive y-direction. Therefore, the x & y coordinates of the final position $(X_{f}, Y_{f})$ are given by the following expressions:

$X_{f}=\frac{2}{2^2}-\frac{4}{2^4}+\frac{6}{2^6}-\frac{8}{2^8}+\cdots$

$Y_{f}=\frac{1}{2}-\frac{3}{2^3}+\frac{5}{2^5}-\frac{7}{2^7}+\cdots$

The following figure shows how the expressions for $X_{f}$ and $Y_{f}$ are derived:

Now, $X_{f}=\frac{1}{2} \displaystyle \sum_{n=1}^{\infty} \frac{n(-1)^{(n-1)}}{2^{2(n-1)}}=\frac{1}{2} \displaystyle \sum_{n=1}^{\infty} \displaystyle \sum_{k=n}^{\infty} \frac{(-1)^{(k-1)}}{2^{2(k-1)}}$

$\Rightarrow X_{f}=\frac{1}{2} \displaystyle \sum_{n=1}^{\infty} [\frac{(-1)^{(n-1)}}{2^{2(n-1)}} \times \frac{1}{1-(-\frac{1}{4})}]=\frac{2}{5} \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{(n-1)}}{2^{2(n-1)}}$

$\Rightarrow X_{f}=\frac{2}{5} \times \frac{1}{1-(-\frac{1}{4})}=\frac{8}{25}$

Next, $Y_{f}=\frac{1}{2} \displaystyle \sum_{n=1}^{\infty} \frac{(2n-1)(-1)^{(n-1)}}{2^{2(n-1)}}=\displaystyle \sum_{n=1}^{\infty} \frac{n(-1)^{(n-1)}}{2^{2(n-1)}}-\frac{1}{2} \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{(n-1)}}{2^{2(n-1)}}$

Note that we have already computed both the sums while calculating $X_{f}$ . The first sum is equal to $\frac{16}{25}$ , and the second sum is equal to $\frac{4}{5}$ . Therefore,

$Y_{f}=\frac{16}{25}-(\frac{1}{2} \times \frac{4}{5})=\frac{6}{25}$ .

Therefore, the net displacement from origin is:

$S_{f}=\sqrt{X_{f}^{2}+Y_{f}^{2}}=\frac{10}{25}=\frac{2}{5}$

Hence, the required answer is $2+5=\boxed{7}$

We see coordinates of the final point are:

$\\ \\ \left( \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\frac { 2n }{ { x }^{ 2n } } ,\quad \quad \quad \quad \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 } } \frac { 2n-1 }{ { x }^{ 2n-1 } } } \right) \\ \therefore \quad Displacement=\quad \zeta \quad =\quad \sqrt { \left( { \left\{ \sum { { (-1) }^{ n+1 }\frac { 2n }{ { x }^{ 2n } } } \right\} }^{ 2 }+{ \left\{ \sum { { (-1) }^{ n+1 }\frac { 2n-1 }{ { x }^{ 2n-1 } } } \right\} }^{ 2 } \right) \\ } \\ Let\quad f\left( x \right) =\quad \sum _{ n=1 }^{ \infty }{ \frac { n }{ { x }^{ n } } } \\ f\left( x \right) -\frac { f\left( x \right) }{ x } =\quad \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { x }^{ n } } } =\quad \frac { 1 }{ x-1 } \Longrightarrow f\left( x \right) =\frac { x }{ { (x-1) }^{ 2 } } \\ Now,\quad -f(ix)=\quad \sum { { (-1) }^{ n+1 }\frac { 2n }{ { x }^{ 2n } } } +\quad i\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 } } \frac { 2n-1 }{ { x }^{ 2n-1 } } =\quad \frac { ix }{ { (ix-1) }^{ 2 } } \\ \therefore \quad \left| f\left( ix \right) \right| =\quad \zeta \quad =\quad \left| \frac { { ix } }{ { (ix-1) }^{ 2 } } \right| =\quad \underset { x=2 }{ \frac { x }{ { x }^{ 2 }+1 } } =\quad \boxed { \frac { 2 }{ 5 } } \\$