Đáp án đúng:
Giải chi tiết:Ta có:
\(\begin{array}{l}S = \frac{5}{{{2^2}}} + \frac{5}{{{3^2}}} + \frac{5}{{{4^2}}} + ... + \frac{5}{{{{100}^2}}}.\\ = 5.\left( {\frac{1}{{2.2}} + \frac{1}{{3.3}} + \frac{1}{{4.4}} + ... + \frac{1}{{100.100}}} \right)\\ > 5.\left( {\frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + ... + \frac{1}{{100.101}}} \right)\\ > 5.\left( {\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{100}} - \frac{1}{{101}}} \right)\\ > 5.\left( {\frac{1}{2} - \frac{1}{{101}}} \right) > \frac{5}{2} > 2\\ \Rightarrow S > 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\end{array}\)
\(\begin{array}{l}S = \frac{5}{{{2^2}}} + \frac{5}{{{3^2}}} + \frac{5}{{{4^2}}} + ... + \frac{5}{{{{100}^2}}}.\\ = 5.\left( {\frac{1}{{2.2}} + \frac{1}{{3.3}} + \frac{1}{{4.4}} + ... + \frac{1}{{100.100}}} \right)\\ < 5.\left( {\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{99.100}}} \right)\\ < 5.\left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{99}} - \frac{1}{{100}}} \right)\\ < 5.\left( {1 - \frac{1}{{100}}} \right) < 5\\ \Rightarrow S < 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\end{array}\)
Từ (1) và (2) : \(2 < S < 5\) (đpcm).
