a) Tính: \(\frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + … + \frac{1}{{19.21}}\). b) Chứng minh: \(A = \frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + … + \frac{1}{{(2n – 1)(2n + 1)}} < \frac{1}{2}\).

a) Tính: \(\frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + … + \frac{1}{{19.21}}\).

b) Chứng minh: \(A = \frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + … + \frac{1}{{(2n – 1)(2n + 1)}} < \frac{1}{2}\).

Phương pháp giải:

+) Áp dụng công thức dạng tổng quát: \(\frac{n}{{m.(m + n)}} = \frac{1}{m} – \frac{1}{{m + n}}\) .

a) Ta có:

\(\begin{array}{l}\;\;\frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + … + \frac{1}{{19.21}}\\ = \frac{1}{2} \cdot \left( {\frac{2}{{1.3}} + \frac{2}{{3.5}} + \frac{2}{{5.7}} + … + \frac{2}{{19.21}}} \right)\\ = \frac{1}{2} \cdot \left( {1 – \frac{1}{3} + \frac{1}{3} – \frac{1}{5} + \frac{1}{5} – \frac{1}{7} + … + \frac{1}{{19}} – \frac{1}{{21}}} \right)\\ = \frac{1}{2} \cdot \left( {1 – \frac{1}{{21}}} \right)\\ = \frac{1}{2} \cdot \frac{{20}}{{21}} = \frac{{10}}{{21}}\end{array}\)

b) Ta có:

\(\begin{array}{l}A = \frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + … + \frac{1}{{(2n – 1)(2n + 1)}}\\\,\,\,\, = \frac{1}{2} \cdot \left( {\frac{2}{{1.3}} + \frac{2}{{3.5}} + \frac{2}{{5.7}} + … + \frac{2}{{(2n – 1)(2n + 1)}}} \right)\\\,\,\,\, = \frac{1}{2} \cdot \left( {1 – \frac{1}{3} + \frac{1}{3} – \frac{1}{5} + … + \frac{1}{{2n – 1}} – \frac{1}{{2n + 1}}} \right)\\\,\,\,\, = \frac{1}{2} \cdot \left( {1 – \frac{1}{{2n + 1}}} \right)\end{array}\)

Lại có \(1 – \frac{1}{{2n + 1}} < 1\) nên suy ra \(A < \frac{1}{2}.1\), hay \(A < \frac{1}{2}\) (đpcm).