Câu hỏi:

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Tự luận

Tính F=118+154+1108+...+1990F = \frac{1}{{18}} + \frac{1}{{54}} + \frac{1}{{108}} + ... + \frac{1}{{990}}.

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Lời giải

Hướng dẫn giải:

Ta có F=118+154+1108+...+1990F = \frac{1}{{18}} + \frac{1}{{54}} + \frac{1}{{108}} + ... + \frac{1}{{990}}

=13.6+16.9+19.12+...+130.33 = \frac{1}{{3.6}} + \frac{1}{{6.9}} + \frac{1}{{9.12}} + ... + \frac{1}{{30.33}}

=13.(33.6+36.9+39.12+...+330.33) = \frac{1}{3}.\left( {\frac{3}{{3.6}} + \frac{3}{{6.9}} + \frac{3}{{9.12}} + ... + \frac{3}{{30.33}}} \right)

=13.(633.6+966.9+1299.12+...+333030.33) = \frac{1}{3}.\left( {\frac{{6 - 3}}{{3.6}} + \frac{{9 - 6}}{{6.9}} + \frac{{12 - 9}}{{9.12}} + ... + \frac{{33 - 30}}{{30.33}}} \right)

=13.[(63.633.6)+(96.966.9)+(129.1299.12)+...+(3330.333030.33)] = \frac{1}{3}.\left[ {\left( {\frac{6}{{3.6}} - \frac{3}{{3.6}}} \right) + \left( {\frac{9}{{6.9}} - \frac{6}{{6.9}}} \right) + \left( {\frac{{12}}{{9.12}} - \frac{9}{{9.12}}} \right) + ... + \left( {\frac{{33}}{{30.33}} - \frac{{30}}{{30.33}}} \right)} \right]

=13.(1316+1619+19112+...+130133) = \frac{1}{3}.\left( {\frac{1}{3} - \frac{1}{6} + \frac{1}{6} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{12}} + ... + \frac{1}{{30}} - \frac{1}{{33}}} \right)

=13.(13133) = \frac{1}{3}.\left( {\frac{1}{3} - \frac{1}{{33}}} \right)=13.1033=1099 = \frac{1}{3}.\frac{{10}}{{33}} = \frac{{10}}{{99}}.

Vậy F=1099F = \frac{{10}}{{99}}.

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