Thực hiện phép tính một cách hợp lí: a) A = (2/3).5+ (2/5). 7 + (2/7). 9 + ... + (2/61).63 b) B =(2/1).4+ (2/4).7+ (2/7 ).11 + ... +(2/100).103

Hướng dẫn giải

a) \(A = \frac{2}{{3 \cdot 5}} + \frac{2}{{5 \cdot 7}} + \frac{2}{{7 \cdot 9}} + ... + \frac{2}{{61 \cdot 63}}\)

\( = \left( {\frac{1}{3} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{7}} \right) + \left( {\frac{1}{7} - \frac{1}{9}} \right) + ... + \left( {\frac{1}{{61}} - \frac{1}{{63}}} \right)\)

\( = \frac{1}{3} - \frac{1}{{63}} = \frac{{21}}{{63}} - \frac{1}{{63}} = \frac{{20}}{{63}}.\)

b) \(B = \frac{2}{{1 \cdot 4}} + \frac{2}{{4 \cdot 7}} + \frac{2}{{7 \cdot 11}} + ... + \frac{2}{{100 \cdot 103}}\)

\( = \frac{2}{3} \cdot \left( {\frac{3}{{1 \cdot 4}} + \frac{3}{{4 \cdot 7}} + \frac{3}{{7 \cdot 11}} + ... + \frac{3}{{100 \cdot 103}}} \right)\)

\( = \frac{2}{3} \cdot \left( {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{11}} + ... + \frac{1}{{100}} - \frac{1}{{103}}} \right)\)

\( = \frac{2}{3} \cdot \left( {1 - \frac{1}{{103}}} \right) = \frac{2}{3} \cdot \frac{{103 - 1}}{{103}}\)

\( = \frac{2}{3} \cdot \frac{{102}}{{103}} = \frac{2}{1} \cdot \frac{{34}}{{103}} = \frac{{68}}{{103}}.\)

c) \(C = \frac{{10}}{{56}} + \frac{{10}}{{140}} + \frac{{10}}{{260}} + ... + \frac{{10}}{{1\,\,400}}\)

\( = \frac{{20}}{{112}} + \frac{{20}}{{280}} + \frac{{20}}{{520}} + ... + \frac{{20}}{{2\,\,800}}\)

\( = \frac{{20}}{6} \cdot \left( {\frac{6}{{8 \cdot 14}} + \frac{6}{{14 \cdot 20}} + \frac{6}{{20 \cdot 26}} + ... + \frac{6}{{50 \cdot 56}}} \right)\)

\( = \frac{{20}}{6} \cdot \left( {\frac{1}{8} - \frac{1}{{14}} + \frac{1}{{14}} - \frac{1}{{20}} + \frac{1}{{20}} - \frac{1}{{26}} + ... + \frac{1}{{50}} - \frac{1}{{56}}} \right)\)

\( = \frac{{10}}{3} \cdot \left( {\frac{1}{8} - \frac{1}{{56}}} \right) = \frac{{10}}{3} \cdot \left( {\frac{7}{{56}} - \frac{1}{{56}}} \right) = \frac{{10}}{3} \cdot \frac{6}{{56}} = \frac{{10}}{3} \cdot \frac{3}{{28}} = \frac{5}{{14}}.\)

d) \(D = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{1\,\,024}}\)

\( = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ...... + \frac{1}{{{2^{10}}}}\)

Đặt \(S = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{10}}}},\) khi đó \(D = 1 + S.\)

Do đó \(2S = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ...... + \frac{1}{{{2^9}}}\)

Suy ra \(2S - S = \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ...... + \frac{1}{{{2^9}}}} \right) - \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ...... + \frac{1}{{{2^9}}} + \frac{1}{{{2^{10}}}}} \right)\)

\(S = 1 - \frac{1}{{{2^{10}}}}\)

Vậy \(D = 1 + S = 1 + 1 - \frac{1}{{{2^{10}}}} = 2 - \frac{1}{{{2^{10}}}}.\)