Infinitely Infinite Resistors

Let the equivalent resistance be $R$ . Since the circuit is infinite, we have (see diagram below):

$\begin{aligned} R & = 1 + 1||2R + 1 \\ & = 2 + \dfrac{1\times 2R}{1+2R} \\ \Rightarrow R(1+2R) & = 2(1 + 2R) + 2R \\ R + 2R^2 & = 2 + 6R \\ 2R^2 - 5R -2 & = 0 \\ \Rightarrow R & = \frac{5+\sqrt{25+16}}{8} = \frac{5+\sqrt{41}}{8} \\ \Rightarrow a + b + c & = 5+41+8 = \boxed{50} \end{aligned}$

The resistance of the circuit can be written as an infinite fraction:

$2({ 2 }^{ 0 })+\frac { 1 }{ \frac { 1 }{ { 2 }^{ 0 } } +\frac { 1 }{ 2({ 2 }^{ 1 })+\frac { 1 }{ \frac { 1 }{ { 2 }^{ 1 } } +\frac { 1 }{ 2({ 2 }^{ 2 })+\frac { 1 }{ ... } } } } } ={ 2 }^{ 1 }+\frac { 1 }{ \frac { 1 }{ { 2 }^{ 0 } } +\frac { 1 }{ { 2 }^{ 2 }+\frac { 1 }{ \frac { 1 }{ { 2 }^{ 1 } } +\frac { 1 }{ { 2 }^{ 3 }+\frac { 1 }{ ... } } } } } =x\quad ---(1)$

And so,

$2x={ 2 }^{ 2 }+\frac { 1 }{ \frac { 1 }{ { 2 }^{ 1 } } +\frac { 1 }{ { 2 }^{ 3 }+\frac { 1 }{ \frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 4 }+\frac { 1 }{ ... } } } } } \quad ---(2)$

This makes:

$(2)\quad substitute\quad into\quad (1):\\ \\ x={ 2 }^{ 1 }+\frac { 1 }{ \frac { 1 }{ { 2 }^{ 0 } } +\frac { 1 }{ { 2 }^{ 2 }+\frac { 1 }{ \frac { 1 }{ { 2 }^{ 1 } } +\frac { 1 }{ { 2 }^{ 3 }+\frac { 1 }{ ... } } } } } ={ 2 }+\frac { 1 }{ 1+\frac { 1 }{ 2x } }$

Solving for:

$x={ 2 }+\frac { 1 }{ 1+\frac { 1 }{ 2x } }$

Gives:

$x=\frac { 5\pm \sqrt { 41 } }{ 4 }$

Since

$x>2$

$x=\frac { 5+\sqrt { 41 } }{ 4 } =\frac { a+\sqrt { b } }{ c }$

Making $a+b+c=\boxed{50}$

For each loop, resistance becomes 2 times. So, if equivalent resistance between A and B is x, then the equivalent resistance of second loop is 2x. Now 2x and 1 in parallel and this is in series with the two 1,s. This is equal to x(equivalent resistance between A and B). This gives us a quadratic in x, (x^2 - 5x - 2 =0). On solving, we get a=5, b=41, c=4. So, a+b+c=50