Phương pháp giải:

Áp dụng câu 14 suy ra \(\sqrt {1 + \frac{1}{{{a^2}}} + \frac{1}{{{{\left( {a + 1} \right)}^2}}}}  = \frac{{{a^2} + a + 1}}{{a\left( {a + 1} \right)}} = 1 + \frac{1}{{a\left( {a + 1} \right)}} = 1 + \frac{1}{a} - \frac{1}{{a + 1}}\)

Lời giải chi tiết:

Áp dụng câu 14 suy ra \(\sqrt {1 + \frac{1}{{{a^2}}} + \frac{1}{{{{\left( {a + 1} \right)}^2}}}}  = \frac{{{a^2} + a + 1}}{{a\left( {a + 1} \right)}} = 1 + \frac{1}{{a\left( {a + 1} \right)}} = 1 + \frac{1}{a} - \frac{1}{{a + 1}}\)

Do đó: \(B = \sqrt {1 + \frac{1}{{{1^2}}} + \frac{1}{{{2^2}}}}  + \sqrt {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}}}  + \sqrt {1 + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}}}  + ... + \sqrt {1 + \frac{1}{{{{99}^2}}} + \frac{1}{{{{100}^2}}}} \)

\(\)\(\begin{array}{l} = \left( {1 + \frac{1}{1} - \frac{1}{2}} \right) + \left( {1 + \frac{1}{2} - \frac{1}{3}} \right) + \left( {1 + \frac{1}{3} - \frac{1}{4}} \right) + ... + \left( {1 + \frac{1}{{99}} - \frac{1}{{100}}} \right)\\ = 99 + \left( {\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{99}} - \frac{1}{{100}}} \right)\\ = 100 - \frac{1}{{100}}\\ = 99,99.\end{array}\)

Chọn D.