23.
Sử dụng: $n^2-1=(n+1)(n-1)$
$B=\dfrac{1^2}{2^2-1}\cdot\dfrac{1^2}{2^2-1}\cdot\dfrac{1^2}{2^2-1}\cdot...\cdot\dfrac{(2n+1)^2}{(2n+2)^2-1}\\=\dfrac{1}{1\cdot3}\cdot\dfrac{3\cdot3}{3\cdot5}\cdot\dfrac{5\cdot5}{5\cdot7}\cdot...\cdot\dfrac{(2n+1)\cdot (2n+1)}{(2n+1)\cdot (2n+3)}\\=\dfrac{1}{3}\cdot \dfrac{3}{5}\cdot \dfrac{5}{7}\cdot ...\cdot\dfrac{2n+1}{2n+3}\\=\dfrac1{2n+3}$.
21.
Sử dụng: $\dfrac{1}{n^2}<\dfrac{1}{n^2-1}=\dfrac{1}{(n-1)(n+1)}$
$A<\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{(2n)\cdot (2n+2)}\\A<\dfrac12\left(\dfrac{1}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{(2n)\cdot (2n+2)}\right)\\A<\dfrac12\left(\dfrac12-\dfrac14+\dfrac14-\dfrac16+\dfrac16-\dfrac18+...+\dfrac1{2n}-\dfrac1{2n+2}\right)\\A<\dfrac12\left(\dfrac12-\dfrac1{2n+2}\right)<\dfrac12\cdot\dfrac12=\dfrac14$
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https://diendan.hocmai.vn/threads/t...c-mon-danh-cho-ban-hoan-toan-mien-phi.827998/
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