Tính tổng sau: \(A = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{99.100}}\)
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A.
\(A = 1\)
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B.
\(A = 0\)
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C.
\(A = \frac{1}{2}\)
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D.
\(A = \frac{{99}}{{100}}\)
Sử dụng công thức \(\frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}\)
\(\begin{array}{l}A = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{99.100}}\\ = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ... + \left( {\frac{1}{{99}} - \frac{1}{{100}}} \right)\\ = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{99}} - \frac{1}{{100}}\\ = 1 - \frac{1}{{100}} = \frac{{99}}{{100}}\end{array}\)
Đáp án : D