Đáp án:
\[C = \frac{{\sqrt {403} - 1}}{{\sqrt 5 }}\]
Giải thích các bước giải:
Tổng quát:
\(\frac{1}{{\sqrt n + \sqrt {n + 1} }} = \frac{{\left( {\sqrt {n + 1} - \sqrt n } \right)}}{{\left( {\sqrt {n + 1} - \sqrt n } \right)\left( {\sqrt {n + 1} + \sqrt n } \right)}} = \frac{{\sqrt {n + 1} - \sqrt n }}{{\left( {n + 1} \right) - n}} = \sqrt {n + 1} - \sqrt n \)
Ta có:
\(\begin{array}{l}
C = \frac{1}{{\sqrt 5 + \sqrt {10} }} + \frac{1}{{\sqrt {10} + \sqrt {15} }} + \frac{1}{{\sqrt {15} + \sqrt {20} }} + ..... + \frac{1}{{\sqrt {2010} + \sqrt {2015} }}\\
= \frac{1}{{\sqrt 5 }}\left( {\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + .... + \frac{1}{{\sqrt {402} + \sqrt {403} }}} \right)\\
= \frac{1}{{\sqrt 5 }}\left( {\sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 + \sqrt 4 - \sqrt 3 + .... + \sqrt {403} - \sqrt {402} } \right)\\
= \frac{1}{{\sqrt 5 }}\left( {\sqrt {403} - 1} \right)\\
= \frac{{\sqrt {403} - 1}}{{\sqrt 5 }}
\end{array}\)
