$\begin{array}{l}1)\\\text{- Đặt $A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}\,+\,...+\,\dfrac{1}{10000}$}\\\to A=\dfrac14\left(1+\dfrac14+\dfrac19+...+\dfrac1{2500}\right)\\\to A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\\\text{- Ta nhận thấy :}\\\dfrac1{2^2}<\dfrac1{1.2}\\\dfrac1{3^2}<\dfrac1{2.3}\\...........\\\dfrac1{50^2}<\dfrac1{49.50}\\\to 1+\dfrac1{2^2}+\dfrac1{3^2}+...+\dfrac1{50^2}<1+\dfrac1{1.2}+\dfrac1{2.3}+...+\dfrac1{49.50}\\\to A< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\\\to A< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\\\to A< \dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)\\\to A< \dfrac{1}{4}.\dfrac{99}{50}\\\to A < \dfrac{99}{200}<\dfrac{100}{200}=\dfrac12\\\to A<\dfrac12\\\to \dfrac14+\dfrac1{16}+\dfrac1{36}+...+\dfrac1{10000}<\dfrac12\\\,\\3)\\\text{- Gọi $d\in ƯC(3n+2,4n+3)\,\,(d \in \mathbb{Z}$)}\\\to \begin{cases} 3n+2\,\,\vdots\,\,d\\4n+3\,\,\vdots\,\,d\end{cases}\\\to \begin{cases} 4(3n+2)\,\,\vdots\,\,d\\3(4n+3)\,\,\vdots\,\,d\end{cases}\\\to \begin{cases} 12n+8\,\,\vdots\,\,d\\12n+9\,\,\vdots\,\,d\end{cases}\\\to (12n+9)-(12n+8)\,\,\vdots\,\,d\\\to 1 \,\,\vdots\,\,d\\\to d=\pm1\\\to\text{$\dfrac{3n+2}{4n+3}$ tối giản} \end{array}$
