Giải hộ mình với ạ giải hộ mình với , mình cảm ơn

Đáp án:

$A = 1980$

Giải thích các bước giải:

$\begin{array}{l}
A = \dfrac{{2000}}{2} + \dfrac{{2000}}{6} + \dfrac{{2000}}{{12}} + ... + \dfrac{{2000}}{{9900}}\\
 = 2000\left( {\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{{12}} + ... + \dfrac{1}{{9900}}} \right)\\
 = 2000\left( {\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{99.100}}} \right)\\
 = 2000\left( {\dfrac{{2 - 1}}{{1.2}} + \dfrac{{3 - 2}}{{2.3}} + \dfrac{{4 - 3}}{{3.4}} + ... + \dfrac{{100 - 99}}{{99.100}}} \right)\\
 = 2000\left( {\dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{99}} - \dfrac{1}{{100}}} \right)\\
 = 2000\left( {1 - \dfrac{1}{{100}}} \right)\\
 = 2000.\dfrac{{99}}{{100}}\\
 = 1980
\end{array}$$\begin{array}{l}
A = \dfrac{{2000}}{2} + \dfrac{{2000}}{6} + \dfrac{{2000}}{{12}} + ... + \dfrac{{2000}}{{9900}}\\
 = 2000\left( {\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{{12}} + ... + \dfrac{1}{{9900}}} \right)\\
 = 2000\left( {\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{99.100}}} \right)\\
 = 2000\left( {\dfrac{{2 - 1}}{{1.2}} + \dfrac{{3 - 2}}{{2.3}} + \dfrac{{4 - 3}}{{3.4}} + ... + \dfrac{{100 - 99}}{{99.100}}} \right)\\
 = 2000\left( {\dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{99}} - \dfrac{1}{{100}}} \right)\\
 = 2000\left( {1 - \dfrac{1}{{100}}} \right)\\
 = 2000.\dfrac{{99}}{{100}}\\
 = 1980
\end{array}$

Vậy $A = 1980$