10000 câu trắc nghiệm tổng hợp môn Toán 2025 mới nhất (có đáp án) - Phần 1

Chứng minh rằng: 1/1.2+1/3.4+1/5.6+...1/49.50=1/26+1/27+1/28+...+1/50

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Chứng minh rằng:

11⋅2+13⋅4+15⋅6+...+149⋅50=126+127+128+...+150.

0/3000 ký tự
Giải thích

\[\frac{1}{{1 \cdot 2}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{5 \cdot 6}} + ... + \frac{1}{{49 \cdot 50}}\]

\[ = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{49}} - \frac{1}{{50}}\]

\[ = \left( {\frac{1}{1} + \frac{1}{3} + ... + \frac{1}{{49}}} \right) - \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{{50}}} \right)\]

\[ = \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{49}} + \frac{1}{{50}}} \right) - \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{{50}}} \right) \cdot 2\]

\[ = \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{49}} + \frac{1}{{50}}} \right) - \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{25}}} \right)\]

\[ = \frac{1}{{26}} + \frac{1}{{27}} + \frac{1}{{28}} + ... + \frac{1}{{50}}\]

Vậy \[\frac{1}{{1 \cdot 2}} + \frac{1}{{3 \cdot 4}} + \frac{1}{{5 \cdot 6}} + ... + \frac{1}{{49 \cdot 50}} = \frac{1}{{26}} + \frac{1}{{27}} + \frac{1}{{28}} + ... + \frac{1}{{50}}.\]