题目详情 - Q20260210222214500

由$a_{n+1}=a_n+\frac{8(n+1)}{(2n+1)^2(2n+3)^2}$，得$a_{n+1}-a_n=\frac{1}{(2n+1)^2}-\frac{1}{(2n+3)^2}$， ∴$a_n-a_{n-1}=\frac{1}{(2n-1)^2}-\frac{1}{(2n+1)^2},\cdots,a_2-a_1=\frac{1}{3^2}-\frac{1}{5^2}$， 则$(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots+(a_2-a_1)$ $=\left(\frac{1}{(2n-1)^2}-\frac{1}{(2n+1)^2}\right)+\left(\frac{1}{(2n-3)^2}-\frac{1}{(2n-1)^2}\right)+\cdots+\left(\frac{1}{3^2}-\frac{1}{5^2}\right)$， 化简得$a_n-a_1=\frac{1}{3^2}-\frac{1}{(2n+1)^2}$，又$a_1=\frac{8}{9}$， 所以$a_n=\frac{(2n+1)^2-1}{(2n+1)^2}$.