Đáp án:
Giải thích các bước giải:
Áp dụng:
\[\begin{array}{l}
1 + 2 + 3 + 4 + .... + n = \frac{{n\left( {n + 1} \right)}}{2}\\
\frac{1}{{n\left( {n + 1} \right)}} = \frac{{\left( {n + 1} \right) - n}}{{n\left( {n + 1} \right)}} = \frac{{n + 1}}{{n\left( {n + 1} \right)}} - \frac{n}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}
\end{array}\]
Ta có:
\[\begin{array}{l}
A = \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}} + .... + \frac{1}{{1 + 2 + 3 + .... + 99}}\\
A = \frac{1}{{\frac{{2.3}}{2}}} + \frac{1}{{\frac{{3.4}}{2}}} + \frac{1}{{\frac{{4.5}}{2}}} + ..... + \frac{1}{{\frac{{99.100}}{2}}}\\
A = 2.\left( {\frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + .... + \frac{1}{{99.100}}} \right)\\
A = 2.\left( {\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + .... + \frac{1}{{99}} - \frac{1}{{100}}} \right)\\
A = 2.\left( {\frac{1}{2} - \frac{1}{{100}}} \right)\\
A = 2.\frac{{49}}{{100}} = \frac{{49}}{{50}}
\end{array}\]
