1/100×99 – 1/99×98 – 1/98×97 – 1/97×96 – 1/96×95 – … – 1/2-1 21/08/2021 Bởi Jasmine 1/100×99 – 1/99×98 – 1/98×97 – 1/97×96 – 1/96×95 – … – 1/2-1
Giải thích các bước giải: Áp dụng: \[\frac{1}{{n\left( {n + 1} \right)}} = \frac{{\left( {n + 1} \right) – n}}{{n\left( {n + 1} \right)}} = \frac{1}{n} – \frac{1}{{n + 1}}\] Ta có: \[\begin{array}{l}P = \frac{1}{{100.99}} – \frac{1}{{99.98}} – \frac{1}{{98.97}} – \frac{1}{{97.96}} – ….. – \frac{1}{{2.1}}\\ \Leftrightarrow P = \frac{1}{{100.99}} – \left( {\frac{1}{{99.98}} + \frac{1}{{98.97}} + \frac{1}{{97.96}} + … + \frac{1}{{2.1}}} \right)\\ \Leftrightarrow P = \frac{1}{{99}} – \frac{1}{{100}} – \left( {\frac{1}{{98}} – \frac{1}{{99}} + \frac{1}{{97}} – \frac{1}{{98}} + \frac{1}{{96}} – \frac{1}{{97}} + … + 1 – \frac{1}{2}} \right)\\ \Leftrightarrow P = \frac{1}{{99}} – \frac{1}{{100}} – \left( {1 – \frac{1}{{99}}} \right)\\ \Rightarrow P = \frac{1}{{99}} – \frac{1}{{100}} – 1 + \frac{1}{{99}} = \frac{2}{{99}} – \frac{{101}}{{100}}\end{array}\] Bình luận
Giải thích các bước giải:
Áp dụng:
\[\frac{1}{{n\left( {n + 1} \right)}} = \frac{{\left( {n + 1} \right) – n}}{{n\left( {n + 1} \right)}} = \frac{1}{n} – \frac{1}{{n + 1}}\]
Ta có:
\[\begin{array}{l}
P = \frac{1}{{100.99}} – \frac{1}{{99.98}} – \frac{1}{{98.97}} – \frac{1}{{97.96}} – ….. – \frac{1}{{2.1}}\\
\Leftrightarrow P = \frac{1}{{100.99}} – \left( {\frac{1}{{99.98}} + \frac{1}{{98.97}} + \frac{1}{{97.96}} + … + \frac{1}{{2.1}}} \right)\\
\Leftrightarrow P = \frac{1}{{99}} – \frac{1}{{100}} – \left( {\frac{1}{{98}} – \frac{1}{{99}} + \frac{1}{{97}} – \frac{1}{{98}} + \frac{1}{{96}} – \frac{1}{{97}} + … + 1 – \frac{1}{2}} \right)\\
\Leftrightarrow P = \frac{1}{{99}} – \frac{1}{{100}} – \left( {1 – \frac{1}{{99}}} \right)\\
\Rightarrow P = \frac{1}{{99}} – \frac{1}{{100}} – 1 + \frac{1}{{99}} = \frac{2}{{99}} – \frac{{101}}{{100}}
\end{array}\]