Cho 

A

- Hướng dẫn giải

Đáp án A

$\begin{array}{l}=>U_n=\frac{(1^2+1)(2^2+1).(3^2+1)(4^2+1)\mathrm{...}\text{[}{(2n-1)}^2+1]\text{[}{\left(2n\right)}^2+1]}{(2^2+1)(3^2+1).(4^2+1)(5^2+1)\mathrm{...}\text{[(}2n{)}^2+1]\text{[}{(2n+1)}^2+1]}\\ =>U_n=\frac{2}{{(2n+1)}^2+1}\end{array}$