If $$\sum_{k=1}^{40} \sqrt{1 + \frac{1}{k^2} + \frac{1}{(k+1)^2}} = a + \frac{b}{c}$$ where $$a, b, c \in \mathbb{N}$$, $$b < c$$, $$\gcd(b,c) = 1$$, then what is the value of $$a+b$$?

The expression under the radical is a perfect square, since $$1 + \frac{1}{k^2} + \frac{1}{(k+1)^2} = \left(1 + \frac{1}{k} - \frac{1}{k+1}\right)^2$$. So each term equals $$1 + \frac{1}{k(k+1)}$$ and the sum telescopes to $$40 + \left(1 - \frac{1}{41}\right) = 40 + \frac{40}{41}$$. Thus $$a = 40$$, $$b = 40$$ and $$a+b = 80$$.