This can't be pretty

The denominators are of the form

$\sqrt{(2k + 1) + 2\sqrt{k(k + 1)}} = \sqrt{(\sqrt{k} + \sqrt{k + 1})^{2}} = \sqrt{k} + \sqrt{k + 1}.$

When rationalized, each term then become

$\dfrac{1}{\sqrt{k} + \sqrt{k + 1}} * \dfrac{\sqrt{k} - \sqrt{k + 1}}{\sqrt{k} - \sqrt{k + 1}} = \sqrt{k + 1} - \sqrt{k}.$

When this is then summed from $k = 1$ to $k = 99$ we get a telescoping sum with the only terms left being

$-\sqrt{1} + \sqrt{100} = \boxed{9}.$

observe that the radical has this pattern: (a+b) + 2sqrt(ab), where a and b are consecutive integers

also, (a+b) + 2sqrt(ab) = (sqrt(a) + sqrt(b))^2

so the expression above is also the same as:

(1/(sqrt(1) + sqrt(2)) + (1/(sqrt(2) + sqrt(3)) + ... + (1/(sqrt(99) + sqrt(100))

rationalize each term!

sqrt(2) - sqrt(1) + sqrt(3) - sqrt(2) + ... + sqrt(99) + sqrt(100)

we are left with: sqrt(100) - sqrt(1) = 10 - 1 = 9

9 is the answer

observe that the radical has this pattern: (a+b) + 2sqrt(ab), where a and b are consecutive integers

also, (a+b) + 2sqrt(ab) = (sqrt(a) + sqrt(b))^2

so the expression above is also the same as:

(1/(sqrt(1) + sqrt(2)) + (1/(sqrt(2) + sqrt(3)) + ... + (1/(sqrt(99) + sqrt(100))

rationalize each term!

sqrt(2) - sqrt(1) + sqrt(3) - sqrt(2) + ... + sqrt(99) + sqrt(100)

we are left with: sqrt(100) - sqrt(1) = 10 - 1 = 9