Cho hai biểu thức sau : A =1/1 nhân 2 + 1/3 nhân 4 + ...+ 1/999 nhân 1 000 và B = 1/(501 nhân1 000) + 1/502 nhân 999 +...+1/999 nhân 502 + 1/1 000 nhân 501. Tính A/B
Hướng dẫn giải
Ta có:
\(A = \frac{1}{{1 \cdot 2}} + \frac{1}{{3 \cdot 4}} + ... + \frac{1}{{999 \cdot 1\,\,000}}\)
\[ = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + ... + \frac{1}{{997}} - \frac{1}{{998}} + \frac{1}{{999}} - \frac{1}{{1\,\,000}}\]
\[ = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + ... + \frac{1}{{997}} + \frac{1}{{998}} + \frac{1}{{999}} + \frac{1}{{1\,\,000}} - 2 \cdot \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{{998}} + \frac{1}{{1\,\,000}}} \right)\]
\[ = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + ... + \frac{1}{{997}} + \frac{1}{{998}} + \frac{1}{{999}} + \frac{1}{{1\,\,000}} - \frac{1}{1} - \frac{1}{2} - \frac{1}{3} - ... - \frac{1}{{499}} - \frac{1}{{500}}\]
\[ = \frac{1}{{501}} + \frac{1}{{502}} + \frac{1}{{503}} + ... + \frac{1}{{1\,\,000}}.\]
Ta có: \[A = \frac{1}{{501}} + \frac{1}{{502}} + \frac{1}{{503}} + ... + \frac{1}{{1\,\,000}}\] và \[A = \frac{1}{{1\,\,000}} + \frac{1}{{999}} + \frac{1}{{998}} + ... + \frac{1}{{502}} + \frac{1}{{501}}\]
Suy ra:
\[2A = \left( {\frac{1}{{501}} + \frac{1}{{1000}}} \right) + \left( {\frac{1}{{502}} + \frac{1}{{999}}} \right) + \left( {\frac{1}{{503}} + \frac{1}{{998}}} \right) + ... + \left( {\frac{1}{{999}} + \frac{1}{{502}}} \right) + \left( {\frac{1}{{1\,\,000}} + \frac{1}{{501}}} \right)\]
\[2A = \frac{{1\,\,501}}{{501 \cdot 1\,\,000}} + \frac{{1\,\,501}}{{502 \cdot 999}} + \frac{{1\,\,501}}{{503 \cdot 998}} + .... + \frac{{1\,\,501}}{{502 \cdot 999}} + \frac{{1\,\,501}}{{501 \cdot 1\,\,000}}\]
\[2A = \frac{{1\,\,501}}{{501 \cdot 1000}} + \frac{{1\,\,501}}{{502 \cdot 999}} + \frac{{1\,\,501}}{{503 \cdot 998}} + ... + \frac{{1\,\,501}}{{502 \cdot 999}} + \frac{{1\,\,501}}{{501 \cdot 1\,\,000}}\]
\[2A = \frac{{1\,\,501}}{{501 \cdot 1000}} + \frac{{1\,\,501}}{{502 \cdot 999}} + \frac{{1\,\,501}}{{503 \cdot 998}} + ... + \frac{{1\,\,501}}{{502 \cdot 999}} + \frac{{1\,\,501}}{{501 \cdot 1\,\,000}}\]
\[2A = 1\,\,501 \cdot \left( {\frac{1}{{501 \cdot 1\,\,000}} + \frac{1}{{502 \cdot 999}} + \frac{1}{{503 \cdot 998}} + ... + \frac{1}{{502 \cdot 999}} + \frac{1}{{501 \cdot 1\,\,000}}} \right)\]
Mà \[B = \frac{1}{{501 \cdot 1\,\,000}} + \frac{1}{{502 \cdot 999}} + ... + \frac{1}{{999 \cdot 502}} + \frac{1}{{1\,\,000 \cdot 501}}\]
Nên \(2A = 1\,\,501B\) nên \(\frac{A}{B} = \frac{{1\,\,501}}{2}.\)