Tính tổng: \(S = \frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + ... + \frac{1}{{19.21}}\) .
Câu 360795: 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. \(S = \frac{-6}{7}\)
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.
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Đáp án : C(0) bình luận (0) lời giải
Giải chi tiết:
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 \(S = \frac{1}{7}\) .
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