If 
$$\sum_{k=1}^{40} \left( \sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}}\right) = a + \frac {b}{c}$$
where $a, b, c \in \mathbb{N}, b < c, gcd(b,c) =1 ,$ then what is the value of $a+ b ?$
1. We start with the given sum:
   \[
   \sum_{k=1}^{40} \left( \sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}}\right)
   \]

2. First, simplify the expression inside the square root:
   \[
   1 + \frac{1}{k^{2}} + \frac{1}{(k+1)^2}
   \]

3. Combine the terms over a common denominator:
   \[
   1 + \frac{1}{k^{2}} + \frac{1}{(k+1)^2} = 1 + \frac{(k+1)^2 + k^2}{k^2(k+1)^2}
   \]

4. Expand and simplify the numerator:
   \[
   (k+1)^2 + k^2 = k^2 + 2k + 1 + k^2 = 2k^2 + 2k + 1
   \]

5. Thus, the expression becomes:
   \[
   1 + \frac{2k^2 + 2k + 1}{k^2(k+1)^2} = \frac{k^2(k+1)^2 + 2k^2 + 2k + 1}{k^2(k+1)^2}
   \]

6. Simplify the numerator:
   \[
   k^2(k+1)^2 + 2k^2 + 2k + 1 = k^4 + 2k^3 + k^2 + 2k^2 + 2k + 1 = k^4 + 2k^3 + 3k^2 + 2k + 1
   \]

7. Therefore:
   \[
   1 + \frac{1}{k^{2}} + \frac{1}{(k+1)^2} = \frac{k^4 + 2k^3 + 3k^2 + 2k + 1}{k^2(k+1)^2}
   \]

8. Notice that:
   \[
   \frac{k^4 + 2k^3 + 3k^2 + 2k + 1}{k^2(k+1)^2} = \left( \frac{k^2 + k + 1}{k(k+1)} \right)^2
   \]

9. Taking the square root:
   \[
   \sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k+1)^2}} = \frac{k^2 + k + 1}{k(k+1)}
   \]

10. Therefore, the sum becomes:
    \[
    \sum_{k=1}^{40} \frac{k^2 + k + 1}{k(k+1)}
    \]

11. Simplify the fraction:
    \[
    \frac{k^2 + k + 1}{k(k+1)} = \frac{k^2}{k(k+1)} + \frac{k}{k(k+1)} + \frac{1}{k(k+1)} = \frac{k}{k+1} + \frac{1}{k} - \frac{1}{k+1}
    \]

12. Thus, the sum becomes:
    \[
    \sum_{k=1}^{40} \left(1 + \frac{1}{k} - \frac{1}{k+1}\right)
    \]

13. Split the sum:
    \[
    \sum_{k=1}^{40} 1 + \sum_{k=1}^{40} \frac{1}{k} - \sum_{k=1}^{40} \frac{1}{k+1}
    \]

14. The first sum is straightforward:
    \[
    \sum_{k=1}^{40} 1 = 40
    \]

15. The second and third sums are telescoping series:
    \[
    \sum_{k=1}^{40} \frac{1}{k} - \sum_{k=1}^{40} \frac{1}{k+1} = \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{40}\right) - \left(\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{41}\right)
    \]

16. Most terms cancel out, leaving:
    \[
    1 - \frac{1}{41} = \frac{40}{41}
    \]

17. Therefore, the sum is:
    \[
    40 + \frac{40}{41}
    \]

18. Express the result in the form \(a + \frac{b}{c}\):
    \[
    40 + \frac{40}{41} = 40 + \frac{40}{41}
    \]

19. Here, \(a = 40\), \(b = 40\), and \(c = 41\). Thus:
    \[
    a + b = 40 + 40 = 80
    \]

The final answer is \(\boxed{80}\)