If ${\sum }_{k=1}^{40}{\left(1+\frac{1}{k^2}+\frac{1}{(k+1{)}^2}\right)}^{1/2}=a+\frac{b}{c}$ where $a,b,c$ are positive integers, $b<c$, and $\text{GCD}(b,c)=1$, then what is the value of $a+b$?

Step 1: Simplify the term inside the square root:

Step 2: Notice that the numerator is a perfect square:

Step 3: Thus, the square root simplifies to:

Step 4: Use partial fractions to rewrite the fractional part:

Step 5: Sum this expression from $k=1$ to $40$:

Step 6: The summation telescopes:

Step 7: Combine the terms:

Step 8: Comparing to $a+\frac{b}{c}$, we get $a=40$, $b=40$, $c=41$. These satisfy $b<c$ and $\text{GCD}(40,41)=1$.

Step 9: Finally, find $a+b$: