ASM 2011 – 2012 Tìm x sao cho: $x + 1/1 x 3 + x + 1/3 x 5 — Không quảng cáo

Phương pháp giải

Áp dụng phương pháp giải bài toán dãy phân số có quy luật.

$\left( {x + \frac{1}{{1 \times 3}}} \right) + \left( {x + \frac{1}{{3 \times 5}}} \right) + \left( {x + \frac{1}{{5 \times 7}}} \right) + .... + \left( {x + \frac{1}{{23 \times 25}}} \right) = 11 \times x + \left( {\frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}}} \right)$

$12 \times x + (\frac{1}{{1 \times 3}} + \frac{1}{{3 \times 5}} + \frac{1}{{5 \times 7}} + .... + \frac{1}{{23 \times 25}}) = 11 \times x + (\frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}})$

$x + \left( {\frac{1}{{1 \times 3}} + \frac{1}{{3 \times 5}} + \frac{1}{{5 \times 7}} + ... + \frac{1}{{23 \times 25}}} \right) = \left( {\frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}}} \right)$    (1)

Ta có $A = \frac{1}{{1 \times 3}} + \frac{1}{{3 \times 5}} + \frac{1}{{5 \times 7}} + ... + \frac{1}{{23 \times 25}}$

$= \frac{1}{2} \times \left( {\frac{2}{{1 \times 3}} + \frac{2}{{3 \times 5}} + \frac{2}{{5 \times 7}} + ... + \frac{2}{{23 \times 25}}} \right)$

$= \frac{1}{2} \times \left( {1 - \frac{1}{{25}}} \right)$

$= \frac{1}{2} \times \frac{{24}}{{25}} = \frac{{12}}{{25}}$

Đặt$B = \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}}$

$3 \times B - B = (1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}}) - (\frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + \frac{1}{{243}}){\text{ }}$

$2 \times B = 1 - \frac{1}{{243}} = \frac{{242}}{{243}}$

$B = \frac{{242}}{{243}}:2 = \frac{{121}}{{243}}$

Thay giá trị của A và B vào (1) ta được:

$x + \frac{{12}}{{25}} = \frac{{121}}{{243}}$

$x = \frac{{109}}{{6075}}$