Đáp án:
$=\frac{1}{100}.\left [\left ( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10} \right)-\left ( \frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right ) \right ]$
Giải thích các bước giải:
$A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\\
=\frac{1}{100}.\left ( \frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110} \right )\\
=\frac{1}{100}.\left ( 1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )\\
=\frac{1}{100}.\left [\left ( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10} \right)-\left ( \frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right ) \right ]$
