Tính A= 1/1+2 + 1/1+2+3 + 1/1+2+3+4 + …+ 1/1+2+3+4+…+99

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1. Đáp án:

   Giải thích các bước giải:

   Áp dụng:
   \[1 + 2 + 3 + … + n = \frac{{n\left( {n + 1} \right)}}{2}\]
   \[\frac{1}{{n\left( {n + 1} \right)}} = \frac{{\left( {n + 1} \right) – n}}{{n\left( {n + 1} \right)}} = \frac{{n + 1}}{{n\left( {n + 1} \right)}} – \frac{n}{{n\left( {n + 1} \right)}} = \frac{1}{n} – \frac{1}{{n + 1}}\]
   Ta có:
   \[\begin{array}{l}
   A = \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}} + …… + \frac{1}{{1 + 2 + 3 + 4 + …. + 99}}\\
   A = \frac{1}{{\frac{{2.3}}{2}}} + \frac{1}{{\frac{{3.4}}{2}}} + \frac{1}{{\frac{{4.5}}{2}}} + ….. + \frac{1}{{\frac{{99.100}}{2}}}\\
   A = 2.\left( {\frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + …. + \frac{1}{{99.100}}} \right)\\
   A = 2.\left( {\frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + …. + \frac{1}{{99}} – \frac{1}{{100}}} \right)\\
   A = 2.\left( {\frac{1}{2} – \frac{1}{{100}}} \right)\\
   A = 2.\frac{{49}}{{100}} = \frac{{49}}{{50}}
   \end{array}\]

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