Đáp án đúng:
Giải chi tiết:Đặt \(A = \frac{1}{4} + \frac{1}{{16}} + \frac{1}{{36}} + \frac{1}{{64}} + \frac{1}{{100}} + \frac{1}{{144}} + \frac{1}{{196}} + .... + \frac{1}{{10000}}\) . Ta cần chứng minh \(A < \frac{1}{2}\).
Ta có:
\(\begin{array}{l}A = \frac{1}{4} + \frac{1}{{16}} + \frac{1}{{36}} + \frac{1}{{64}} + \frac{1}{{100}} + \frac{1}{{144}} + \frac{1}{{196}} + .... + \frac{1}{{10000}}\\\,\,\,\,\, = \frac{1}{{2.2}} + \frac{1}{{4.4}} + \frac{1}{{6.6}} + \frac{1}{{8.8}} + \frac{1}{{10.10}} + \frac{1}{{12.12}} + \frac{1}{{14.14}} + ... + \frac{1}{{100.100}}\\\,\,\,\,\, = \frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{8^2}}} + \frac{1}{{{{10}^2}}} + ... + \frac{1}{{{{100}^2}}}\\2A = \frac{2}{{2.2}} + \frac{2}{{4.4}} + \frac{2}{{6.6}} + \frac{2}{{8.8}} + .... + \frac{2}{{100.100}}\\\,\,\,\,\,\,\, < \frac{1}{2} + \frac{2}{{2.4}} + \frac{2}{{4.6}} + \frac{2}{{6.8}} + \frac{2}{{8.10}} + ... + \frac{2}{{98.100}}\\\,\,\,\,\,\,\, < \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{6} + \frac{1}{6} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{10}} + ... + \frac{1}{{98}} - \frac{1}{{100}}\\\,\,\,\,\,\,\, < 1 - \frac{1}{{100}}\\ \Rightarrow 2A < 1 - \frac{1}{{100}}\\ \Rightarrow \,\,\,\,A < \frac{1}{2} - \frac{1}{{200}}\\ \Rightarrow \,\,\,\,\,A < \frac{1}{2}\end{array}\)
Vậy \(A < \frac{1}{2}\) (đpcm).
