Are you good with sum?

Solving the first sum will give f(x) = ${ x }^{ 2 }-15x+54$ .

For g(x) we are going to split the function in parts. First ${ e }^{ \pi i }$ is Euler's Identity and results in -1.

$\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { n }{ { 2 }^{ n } } }$ <- To solve this sum start by writing the first terms. This will give $\frac { 1 }{ 2 } +\frac { 2 }{ 4 } +\frac { 3 }{ 8 } +\frac { 4 }{ 16 } +\frac { 5 }{ 32 } ...$ . We will start by removing $\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { 2 }^{ n } } }$ which can be easily associated with an infinite geometric geometric progression starting at $\frac { 1 }{ 2 }$ and multiplying each term by $\frac { 1 }{ 2 }$ . You can solve that by $\frac { { _{ }{ A }_{ 1 } } }{ 1-r }$ where A1 is the first term and r is the ratio. That results in 1. Coming back to the sum, removing 1 will result in $\frac { 1 }{ 4 } +\frac { 2 }{ 8 } +\frac { 3 }{ 16 } +\frac { 4 }{ 32 } ...$ Now we can remove the same sequence, except that this time we will start with $\frac { 1 }{ 4 }$ . As you might expect, this will result in 1-0.5 = 0.5. Note that we will always do the same procedure starting by the next term. As a result we will get a sequence that is $1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\frac { 1 }{ 16 } +\frac { 1 }{ 32 } ...$ . This sequence equals 2.

${ 2 }^{ 2 }$ = 4. 4! = 24. 24-1 = 23.

For the last part $\displaystyle \sum _{ j=1 }^{ \infty }{ \frac { 1 }{ { \left( -1 \right) }^{ j+1 }{ 2 }^{ j } } }$ will give us $\frac { 1 }{ 2 } -\frac { 1 }{ 4 } +\frac { 1 }{ 8 } -\frac { 1 }{ 16 } +\frac { 1 }{ 32 } ...$ This can be split into 2 geometric progressions. The first starts in $\frac { 1 }{ 2 }$ and has the ratio $\frac { 1 }{ 4 }$ . The second starts at $\frac { 1 }{ 4 }$ and its ratio is also $\frac { 1 }{ 4 }$ . This leads us to $\frac { \frac { 1 }{ 2 } }{ 1-\frac { 1 }{ 4 } } -\frac { \frac { 1 }{ 4 } }{ 1-\frac { 1 }{ 4 } }$ = $\frac { 1 }{ 3 }$ .

$\left\lfloor \frac { 400 }{ \frac { 1 }{ \frac { 1 }{ 3 } } } \right\rfloor$ = 133

We are left with $g\left( x \right) =-{ x }^{ 2 }+23x-133$ Now the hard part...

Note that for our two parabolas, since that g(x) is mirroring f(x) and is a bit translated, we can assume that in the shortest point of both, the tangent will be equal. Let's take a deeper look at that statement.

f(x) = ${ x }^{ 2 }-15x+54$ .

f'(x) = 2x-15

In the second equation I'm going to switch the variables so we don't get confused $g\left( p \right) =-{ p }^{ 2 }+23p-133$

g'(p) = -2p+23

Then, if both tangents must be equal:

2x-15 = -2p+23

$\boxed { x=-p+19 }$

With this result we are going to analise the f(-p+19), which is equal to ${ p }^{ 2 }-23p+130$

The distance between 2 points can be described by $\sqrt { { \left( x-p \right) }^{ 2 }+{ \left( f\left( x \right) -g\left( x \right) \right) }^{ 2 } }$

We can arrange this formula as $\sqrt { { \left( -p+19-\left( p \right) \right) }^{ 2 }+{ \left( { p }^{ 2 }-23p+130-\left( -{ p }^{ 2 }+23p-133 \right) \right) }^{ 2 } }$

Squaring both sides we have that the distance (d) is: ${ d }^{ 2 }={ \left( -2p+19 \right) }^{ 2 }+{ \left( { -2p }^{ 2 }-46p+266 \right) }^{ 2 }$

It is important to keep track of the meaning of the equations we are getting. This one gives the distance between two points with parallel tangents. We want the shortest right? Differentiate the equation to get to $8(2{ p }^{ 3 }-69{ p }^{ 2 }+793p-3034)\quad =\quad 0$ .

Input this equation into Wolfram Alpha and it will give you $\boxed { p=10.417 }$

Put this value in the first equation of distance to get to $\boxed { 2.01963 }$ . It approximate fraction is $\frac { 103 }{ 51 }$

A+B = $\boxed { 154 }$