Ta có dãy số :

\(1\text{ };\text{ }6\text{ };\text{ }11\text{ };\text{ }21\text{ };\text{ }........\)

\(1-1=0\text{ }⋮\text{ }2\)

\(11-1=10\text{ }⋮\text{ }2\)

\(21-1=20\text{ }⋮\text{ }2\)

\(2000-1=1999\text{ }⋮̸2\)

2000 không thuộc dãy số trên .

\(14\text{ }⋮\text{ }2n-1\)

\(\Rightarrow2n-1\inƯ\left(14\right)\)

\(Ư\left(14\right)=\left\{\pm1\text{ };\text{ }\pm2\text{ };\text{ }\pm7\text{ };\text{ }\pm14\right\}\)

Mà 2n - 1 lẻ nên :

\(2n-1\in\left\{\pm1\text{ };\text{ }\pm7\right\}\)

\(n\in\left\{1\text{ };\text{ }0\text{ };\text{ }4\text{ };\text{ }-3\right\}\)

\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)

\(=1-\dfrac{1}{46}< 1\)

Vậy S < 1 ( ĐPCM )

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{240}\right)\)

\(A=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+....+\dfrac{1}{15\times16}\right)\)

\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)

Vì \(\overline{a3b}\) \(=\dfrac{3}{4}\cdot\overline{3ab}\)

\(\Rightarrow\overline{a3b}=\overline{3ab}\cdot\dfrac{3}{4}\)

\(\Rightarrow\overline{10ab}+3=\left(300+\overline{ab}\right)\cdot\dfrac{3}{4}\)

\(\Rightarrow\overline{10ab}+3=225+\dfrac{3}{4}\cdot\overline{ab}\)

\(\Rightarrow\overline{10ab}-\dfrac{3}{4}\cdot\overline{ab}=225-3\)

\(=>\dfrac{37}{4}\cdot\overline{ab}=222\)

\(\Rightarrow\overline{ab}=222\text{ }:\text{ }\dfrac{37}{4}=24\)

Vậy số cần tìm là 24