by Tính tổng: \(S = \frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + … + \frac{1}{{19.21}}\) .
A. \(S = \frac{-1}{7}\)
B. \(S = \frac{6}{7}\)
C. \(S = \frac{1}{7}\)
D. \(S = \frac{-6}{7}\)
Hướng dẫn
Chọn đáp án là: C
Phương pháp giải:
Đánh giá biểu thức, ta thấy, ở mỗi mẫu, mỗi thừa số hơn kém nhau 2 đơn vị. Từ đó biến đổi
\(\begin{array}{l}\frac{1}{{3.5}} = \frac{1}{2}.\frac{{5 – 3}}{{3.5}} = \frac{1}{2}.\left( {\frac{1}{3} – \frac{1}{5}} \right)\\\frac{1}{{5.7}} = \frac{1}{2}.\frac{{7 – 5}}{{5.7}} = \frac{1}{2}.\left( {\frac{1}{5} – \frac{1}{7}} \right)\\……..\\\frac{1}{{19.21}} = \frac{1}{2}.\frac{{21 – 19}}{{19.21}} = \frac{1}{2}.\left( {\frac{1}{{19}} – \frac{1}{{21}}} \right)\end{array}\)
Sau đó cộng vế với vế, ta dễ dàng tính tổng của S.
Ta có:
\(\begin{array}{l}S = \frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + … + \frac{1}{{19.21}} = \frac{1}{2}.\left( {\frac{{5 – 3}}{{3.5}} + \frac{{7 – 5}}{{5.7}} + \frac{{9 – 7}}{{7.9}} + … + \frac{{21 – 19}}{{19.21}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}.\left( {\frac{1}{3} – \frac{1}{5} + \frac{1}{5} – \frac{1}{7} + \frac{1}{7} – \frac{1}{9} + … + \frac{1}{{19}} – \frac{1}{{21}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}.\left( {\frac{1}{3} – \frac{1}{{21}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}.\left( {\frac{7}{{21}} – \frac{1}{{21}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}.\frac{2}{7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{7}\end{array}\)
Vậy \(S = \frac{1}{7}\) .
Chọn C
