Đặt f(n) = (n^2 +n+1)^2. xét dãy số (un) sao cho

* Hướng dẫn giải

Đáp án là D.

Ta có

$$\begin{gathered} f\left(n\right)={\left[\left(n^2+1\right)+n\right]}^2+1 \\ ={\left(n^2+1\right)}^2+2n.\left(n^2+1\right)+n^2+1 \\ =\left(n^2+1\right)\left[n^2+1+2n+1\right] \end{gathered}$$

$=\left(n^2+1\right)\left[{\left(n+1\right)}^2+1\right]$

Do đó:$\frac{f\left(2n-1\right)}{f\left(2n\right)}=\frac{\left[{\left(2n-1\right)}^2+1\right]\left[{\left(2n\right)}^2+1\right]}{\left[{\left(2n\right)}^2+1\right]\left[{\left(2n+1\right)}^2+1\right]}=\frac{{\left(2n-1\right)}^2+1}{{\left(2n+1\right)}^2+1}$

Suy ra

 $$\begin{gathered} u_n=\frac{f\left(1\right).f\left(3\right).f\left(5\right)\mathrm{...}f\left(2n-1\right)}{f\left(2\right).f\left(4\right).f\left(6\right)\mathrm{...}f\left(2n\right)} \\ =\frac{f\left(1\right)}{f\left(2\right)}\cdot \frac{f\left(3\right)}{f\left(4\right)}\cdot \frac{f\left(5\right)}{f\left(6\right)}\cdot \cdot \cdot \frac{f\left(2n-1\right)}{f\left(2n\right)} \end{gathered}$$

$$\begin{gathered} =\frac{1^2+1}{3^2+1}\cdot \frac{3^2+1}{5^2+1}\cdot \frac{5^2+1}{7^2+1}\cdot \cdot \cdot \frac{{\left(2n-1\right)}^2+1}{{\left(2n+1\right)}^2+1} \\ =\frac{2}{{\left(2n+1\right)}^2+1}=\frac{1}{2n^2+2n+1} \end{gathered}$$

$\Rightarrow n\sqrt{u_n}=n.\sqrt{\frac{1}{2n^2+2n+1}}$

$\Rightarrow \lim n\sqrt{u_n}=\frac{1}{\sqrt{2}}$