Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
E = \frac{1}{{1.101}} + \frac{1}{{2.102}} + \frac{1}{{3.103}} + .... + \frac{1}{{10.110}}\\
\Leftrightarrow 100E = \frac{{100}}{{1.101}} + \frac{{100}}{{2.102}} + \frac{{100}}{{3.103}} + .... + \frac{{100}}{{10.110}}\\
\Leftrightarrow 100E = \frac{{101 - 1}}{{1.101}} + \frac{{102 - 2}}{{2.102}} + \frac{{103 - 3}}{{3.103}} + .... + \frac{{110 - 10}}{{10.110}}\\
\Leftrightarrow 100E = 1 - \frac{1}{{101}} + \frac{1}{2} - \frac{1}{{102}} + \frac{1}{3} - \frac{1}{{103}} + .... + \frac{1}{{10}} - \frac{1}{{110}}\\
\Leftrightarrow E = \frac{1}{{100}}.\left[ {\left( {1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{{10}}} \right) - \left( {\frac{1}{{101}} + \frac{1}{{102}} + .... + \frac{1}{{110}}} \right)} \right]\\
F = \frac{1}{{1.11}} + \frac{1}{{2.12}} + \frac{1}{{3.13}} + ... + \frac{1}{{100.110}}\\
\Leftrightarrow 10F = \frac{{10}}{{1.11}} + \frac{{10}}{{2.12}} + \frac{{10}}{{3.13}} + ... + \frac{{10}}{{100.110}}\\
\Leftrightarrow 10F = \frac{{11 - 1}}{{1.11}} + \frac{{12 - 2}}{{2.12}} + \frac{{13 - 3}}{{3.13}} + ... + \frac{{110 - 100}}{{100.110}}\\
\Leftrightarrow 10F = 1 - \frac{1}{{11}} + \frac{1}{2} - \frac{1}{{12}} + \frac{1}{3} - \frac{1}{{13}} + .... + \frac{1}{{100}} - \frac{1}{{110}}\\
\Leftrightarrow 10F = \left( {1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{{100}}} \right) - \left( {\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + .... + \frac{1}{{110}}} \right)\\
\Leftrightarrow F = \frac{1}{{10}}.\left[ {\left( {1 + \frac{1}{2} + .... + \frac{1}{{10}}} \right) - \left( {\frac{1}{{101}} + \frac{1}{{102}} + ... + \frac{1}{{110}}} \right)} \right]\\
\Rightarrow \frac{E}{F} = \frac{1}{{10}}
\end{array}\)
