Let The Telescoping Dance

Observing the series --
$\sqrt{ 1 + \frac{1}{1} + \frac{1}{4} } = \frac{3}{2}$
$\sqrt{ 1 + \frac{1}{4} + \frac{1}{9} } = \frac{7}{6}$
$\sqrt{ 1 + \frac{1}{9} + \frac{1}{16} } = \frac{13}{12}$
$\sqrt{ 1 + \frac{1}{16} + \frac{1}{25} } = \frac{21}{20}$

So the general term can be put as -
$t_{n} = \frac{ (n+1)^{2} - n }{ (n+1)^{2} - (n+1) } = \frac{ n^2 + n + 1 }{n^2 + n }$
$= 1 + \frac{1}{n(n+1)} = 1 + \frac{1}{n} - \frac{1}{n+1}$

Sum till $n=2013$
$= 1*2013 + \frac{1}{1} - \frac{1}{2014} = \frac{(2013) * (2015)}{2014} = \frac{ac}{b}$

The summation is given as follows, where $n=2013$ :

$\begin{aligned} S & = \sum_{k=1}^n \sqrt{1+\frac 1{k^2} + \frac 1{(k+1)^2}} \\ & = \sum_{k=1}^n \sqrt{\frac {k^2(k+1)^2 + (k+1)^2 + k^2}{k^2(k+1)^2}} \\ & = \sum_{k=1}^n \sqrt{\frac {k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}} \\ & = \sum_{k=1}^n \sqrt{\frac {(k^2+k+1)^2}{k^2(k+1)^2}} \\ & = \sum_{k=1}^n \frac {k^2+k+1}{k(k+1)} \\ & = \sum_{k=1}^n \left(1 + \frac 1{k(k+1)} \right) \\ & = \sum_{k=1}^n \left(1 + \frac 1k - \frac 1{k+1} \right) \\ & = n + \frac 11 - \frac 1{n+1} \\ & = \frac {(n+1)^2-1}{n+1} \\ & = \frac {n(n+2)}{n+1} \end{aligned}$

$\implies a+b+c = n+n+1+n+2 = 3(n+1) = 3(2014) = \boxed{6042}$